mirror of
https://github.com/CaiJimmy/hugo-theme-stack.git
synced 2025-04-29 03:53:30 +08:00
986 lines
22 KiB
Markdown
986 lines
22 KiB
Markdown
+++
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author = "Wxn"
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title = "给个offer"
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date = "2024-04-24"
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description = "Please read me first."
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tags = [
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"Dilay",
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]
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categories = [
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"面试复盘",
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]
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+++
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This article offers a sample of basic Markdown.
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<!--more-->
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# 正文开始
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如果考到数论的题,会做你就做,不会做你就说你对数论这块不太熟悉,数论这个面算法工程师考的比较多,做开发的话链表比较多
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## 1.41. 包含min函数的栈
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```cpp
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//https://leetcode.cn/problems/bao-han-minhan-shu-de-zhan-lcof/description/
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class MinStack {
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public:
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/** initialize your data structure here. */
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//1.单调栈
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//2.主栈与辅助栈
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//3.1)push都要插入（如果辅助栈为空，或者辅助栈顶>=x，则辅助栈插入）
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//2)pop() 如果辅助栈顶 == 主栈顶，则辅助栈顶弹出，否者就只有主栈弹出
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//3)4)直接返回相应的栈
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MinStack() {
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}
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void push(int x) {
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}
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void pop() {
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}
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int top() {
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}
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int getMin() {
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}
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};
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/**
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* Your MinStack object will be instantiated and called as such:
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* MinStack obj = new MinStack();
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* obj.push(x);
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* obj.pop();
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* int param_3 = obj.top();
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* int param_4 = obj.getMin();
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*/
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```
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## 2.35. 反转链表
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```cpp
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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//https://leetcode.cn/problems/reverse-linked-list/
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//难点：单链表要建立一个前驱节点
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//关键点：画图
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class Solution {
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public:
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ListNode* reverseList(ListNode* head) {
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}
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};
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```
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## 3.19. 二叉树的下一个节点
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode *father;
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* TreeNode(int x) : val(x), left(NULL), right(NULL), father(NULL) {}
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* };
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*/
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//这个是vip题目
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//285. 二叉搜索树中的中序后继
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//分类讨论:有右子树和没有右子树
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//中序遍历
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//情况1:这个点有右子树,那后继就是"右子树"最左边的那个
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//情况2:这个点的右子树为空(且有父节点),当p有父节点且p等于p父节点的右儿子,那么p就赋值为p的父节点,最后返回p的父节点
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class Solution {
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public:
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TreeNode* inorderSuccessor(TreeNode* p) {
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}
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};
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```
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## 4.34. 链表中环的入口结点
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[142. 环形链表 II](https://leetcode.cn/problems/linked-list-cycle-ii/)
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[LCR 022. 环形链表 II](https://leetcode.cn/problems/c32eOV/)
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```cpp
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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ListNode *entryNodeOfLoop(ListNode *head) {
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}
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};
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```
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画图:数学证明:(b+c)表示n圈
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```cpp
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备份
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class Solution {
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public:
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ListNode *entryNodeOfLoop(ListNode *head) {
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auto first = head,slow = head;
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while(first && first->next && first->next->next)
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{
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first = first->next->next;
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slow = slow->next;
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if(first == slow)
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{
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first = head;
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while(first != slow)
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{
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first = first->next;
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slow = slow->next;
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}
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return first;
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}
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}
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return nullptr;
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}
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};
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```
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## 5.77.翻转单词顺序
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[151. 反转字符串中的单词](https://leetcode.cn/problems/reverse-words-in-a-string/)
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```cpp
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//先翻转整个句子
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//再翻转单独的一个单词
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//难点:在找到一段时,不要忘记边界
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```
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```cpp
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void Reverse(int l ,int r,string& s)
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{
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for(int i = l , j = r ;i < j ;i++,j--)swap(s[i],s[j]);
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}
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Reverse(0,s.size()-1,s);
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等价于
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reverse(s.begin()+0,s.begin()+s.size());//范围:[)
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反转不是空格的那一段
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```
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## 6.18.重建二叉树
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https://leetcode.cn/problems/zhong-jian-er-cha-shu-lcof/description/
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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//前:根左右
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//中:左根右
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//1.使用哈希表,快速的找到"一个元素在中序遍历的位置"
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//2.递归dfs(主函数直接返回)
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//1)递归参数:左右子树节点个数
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//2)递归内部:
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/*
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- 前序遍历:左>右 -> null
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- 根节点的值为前序遍历的第1个点 preorder[a]
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- 找到根节点在哈希表中的位置
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- 左右子树递归创建 范围画图
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*/
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class Solution {
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public:
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TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
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}
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};
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```
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||
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||
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||
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||
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||
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||
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## 7.21. 斐波那契数列
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```cpp
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//f[i] = f[i-1]+f[i-2]
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class Solution {
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public:
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int Fibonacci(int n) {
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}
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};
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```
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## 8.78. 左旋转字符串
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```cpp
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//先把整个进行翻转
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//再把前(总-个数),后两部分进行翻转
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class Solution {
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public:
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string leftRotateString(string str, int n) {
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}
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};
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```
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## 9.87. 把字符串转换成整数
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```cpp
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//分步
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//过滤掉行首空格
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//long long
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//判断这个数是不是负数
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//如果在累加的过程中(还没加完),就已经越界了,那就直接跳出来
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class Solution {
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||
public:
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int strToInt(string str) {
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}
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};
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```
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## 10.28. 在O(1)时间删除链表结点
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|
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```cpp
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/**
|
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* Definition for singly-linked list.
|
||
* struct ListNode {
|
||
* int val;
|
||
* ListNode *next;
|
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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//1.用下一个节点覆盖掉当前节点
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//2.删除掉当前节点
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class Solution {
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||
public:
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void deleteNode(ListNode* node) {
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}
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};
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```
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||
|
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## 11.66. 两个链表的第一个公共结点
|
||
|
||
```cpp
|
||
/**
|
||
* Definition for singly-linked list.
|
||
* struct ListNode {
|
||
* int val;
|
||
* ListNode *next;
|
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* ListNode(int x) : val(x), next(NULL) {}
|
||
* };
|
||
*/
|
||
//双指针,
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//指针1走完a再走b,指针2走完b再走a,返回最后相遇的位置,如果最后都指向空也算相遇了
|
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//这个也是有难点的: 注意:对于指针1与指针2,要么是"回头",要么是"下一个"
|
||
class Solution {
|
||
public:
|
||
ListNode *findFirstCommonNode(ListNode *headA, ListNode *headB) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
|
||
|
||
|
||
|
||
|
||
|
||
## 12.84. 求1+2+…+n
|
||
|
||
```cpp
|
||
//语法题:(false && 条件); = false 可以起到if的效果
|
||
class Solution {
|
||
public:
|
||
int getSum(int n) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
## 13.36. 合并两个排序的链表
|
||
|
||
```cpp
|
||
/**
|
||
* Definition for singly-linked list.
|
||
* struct ListNode {
|
||
* int val;
|
||
* ListNode *next;
|
||
* ListNode(int x) : val(x), next(NULL) {}
|
||
* };
|
||
*/
|
||
//归并排序(一个函数足够)
|
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//虚拟节点(比如说初始化为-1)+当前节点
|
||
//1)两个指针
|
||
//2)比较两个指针的值哪个小,哪个小就给哪个(如果l1)
|
||
//3)链到虚拟链表,当前节点后移,l1也后移
|
||
//4)处理残局:l1与l2哪个不同,就一直链到空为止
|
||
class Solution {
|
||
public:
|
||
ListNode* merge(ListNode* l1, ListNode* l2) {
|
||
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||
}
|
||
};
|
||
```
|
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## 14.14. 不修改数组找出重复的数字
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||
|
||
```cpp
|
||
//简单方法:哈希表
|
||
//方二:抽屉原理
|
||
//二分法:
|
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//一个萝卜一个坑,假如说数的个数>坑的个数,那么这个区间一定存在重复的数
|
||
//注意:给定一个长度为 n+1的数组nums，数组中所有的数均在 1∼n的范围内，其中 n≥1,表示下标有效的范围是[1,nums.size()-1]
|
||
class Solution {
|
||
public:
|
||
int duplicateInArray(vector<int>& nums) {
|
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// l 和 r 分别代表的是 数字 1 和 数字n 这里并不是下标.
|
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int l = 1, r = nums.size() - 1;
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while (l < r){
|
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// 二分 找到中间的那个数
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int mid = l + r >> 1;
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||
int s = 0;
|
||
// 下面这句话的意思 从 nums里面 循环去先去 判断 这个数 x 的值看他是否在 [l, mid]中间, 在的话 判断条件执行完为true
|
||
// true 的话代表 数字 1 flase 代表 数字 0. 然后再进行累加 s += x 统计符合条件的个数.
|
||
// 最终的效果就是 统计了 整个数组中 数的值 在 [l,mid] 之间的个数.
|
||
for (auto x : nums) s += x >= l && x <= mid; // left : [l, mid] , right : [mid + 1, r]
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// 理解: 一个坑存一个数, 正常情况下 一定是坑的个数 和 数的个数相等. 如果一坑里面有两个数. 那么就会出现
|
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// 数的个数 大于 坑的个数 说明 这个区间段一定存在重复的个数.
|
||
if (s > mid - l + 1) r = mid;
|
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else l = mid + 1;
|
||
}
|
||
|
||
return r;//l和r都可以
|
||
}
|
||
};
|
||
```
|
||
|
||
## 15.68. 0到n-1中缺失的数字
|
||
|
||
```cpp
|
||
//二分
|
||
//通过下标直接查找,
|
||
//因为是连续的,如果不少的话,则有nums[mid] == mid,如果是左边少了,就是不等
|
||
//最后还要判断nums[r] == r,要是等于的话,那就是少最后一个
|
||
class Solution {
|
||
public:
|
||
int getMissingNumber(vector<int>& nums) {
|
||
|
||
}
|
||
};
|
||
|
||
|
||
eg:
|
||
//[0,1,2,4] n=4
|
||
//[0,1,2,3] n=4
|
||
class Solution {
|
||
public:
|
||
int getMissingNumber(vector<int>& nums) {
|
||
if(nums.empty())return 0;
|
||
int n = nums.size();
|
||
if(nums.back() == n-1)return n;//缺少最后面一个数
|
||
|
||
int l = 0 ,r = n-1;
|
||
while(l < r)
|
||
{
|
||
int mid = l + r >> 1;
|
||
if(nums[mid] == mid)l = mid +1;//这部分不缺
|
||
else r = mid;
|
||
}
|
||
return r;
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
### 补充：13. 找出数组中重复的数字
|
||
|
||
```cpp
|
||
//哈希表秒了
|
||
//方二：不要了
|
||
class Solution {
|
||
public:
|
||
int duplicateInArray(vector<int>& nums) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 16.75. 和为S的两个数字
|
||
|
||
```cpp
|
||
//时间复杂度最重要
|
||
//用法哈希表,count看是否存在
|
||
class Solution {
|
||
public:
|
||
vector<int> findNumbersWithSum(vector<int>& nums, int target) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
## 17.23. 矩阵中的路径
|
||
|
||
```cpp
|
||
//dfs
|
||
//枚举起点，枚举方向
|
||
//起点怎么枚举:两个for循环
|
||
//方向怎么枚举?上右下左:画图
|
||
|
||
//dfs中:
|
||
//1)刚好遍历到的字符个数u == 字符串的长度,就OK
|
||
//2)如果u下标字符 != 遍历到的字符,就不ok
|
||
//3)遍历四个方向,为了避免回头遍历,比如你刚遍历完一个字符'a',下一个字符还是'a',就会回头遍历,
|
||
//我们要避免这个,就需要先修改为一个其他的,等遍历完再返还
|
||
class Solution {
|
||
public:
|
||
bool dfs(vector<vector<char>>& matrix, string &str,int u ,int x,int y)
|
||
{
|
||
if(str[u] != matrix[x][y])return false;//当前字符与之不符
|
||
if(u == str.size()-1)return true;//刚好是最后一个,而且能来到这,说明相符
|
||
int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
|
||
char t = matrix[x][y];
|
||
matrix[x][y] = '*';
|
||
for(int i = 0 ; i < 4 ;i++)
|
||
{
|
||
int a= x+dx[i],b = y +dy[i];
|
||
if(a>=0 && a<matrix.size() &&b>=0 && b<matrix[a].size())
|
||
if(dfs(matrix,str,u+1,a,b))return true;
|
||
}
|
||
matrix[x][y] = t;
|
||
return false;
|
||
}
|
||
bool hasPath(vector<vector<char>>& matrix, string &str) {
|
||
for(int i = 0 ; i < matrix.size() ; i++)
|
||
{
|
||
for(int j = 0 ; j < matrix[i].size() ; j++)
|
||
{
|
||
if(dfs(matrix,str,0,i,j))
|
||
{
|
||
return true;
|
||
}
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 18.55. 连续子数组的最大和
|
||
|
||
```cpp
|
||
s表示收益
|
||
res表示最终结果
|
||
```
|
||
|
||
|
||
|
||
## 19.42. 栈的压入、弹出序列
|
||
|
||
```cpp
|
||
关键:不是所有的数都入栈,他才开始进行弹出操作.
|
||
完全可以实现这么一种情况:你还进去,人家就已经出来了
|
||
|
||
//1.长度不同,肯定不行
|
||
//2.用一个栈来模拟整个过程
|
||
//3.一直将pushV的元素入栈,直到栈顶元素 == 要弹出的第i元素(i从0开始)
|
||
//(while循环 --->尽可能的把能弹出的元素弹出来[前提:栈中有元素且栈顶元素刚好是我们要弹出的元素])
|
||
//4.最后如果栈为空,即为ok
|
||
class Solution {
|
||
public:
|
||
bool isPopOrder(vector<int> pushV,vector<int> popV) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 20.70. 二叉搜索树的第k个结点
|
||
|
||
[230. 二叉搜索树中第K小的元素](https://leetcode.cn/problems/kth-smallest-element-in-a-bst/)
|
||
|
||
```cpp
|
||
//例子中:第1小的数是1;第二小数是2;第三小的数是3;
|
||
//所有:就是求中序遍历的第k个
|
||
//中序遍历:我们都知道是左根右,左和右就是一个dfs,那么中是难点
|
||
//中序遍历的中:k--,如果k==0,就是我们的答案
|
||
//最后dfs加一个阀门:如果节点指针为空,则直接return
|
||
/**
|
||
* Definition for a binary tree node.
|
||
* struct TreeNode {
|
||
* int val;
|
||
* TreeNode *left;
|
||
* TreeNode *right;
|
||
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
|
||
* };
|
||
*/
|
||
class Solution {
|
||
public:
|
||
TreeNode* kthNode(TreeNode* root, int k) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 21.48. 复杂链表的复刻
|
||
|
||
```cpp
|
||
//1.给旧链表每2个节点之间加1个节点(新节点的值是前节点的值)
|
||
//2.重新遍历链表中的每个节点 (p->next)->random=(p->random)->next;
|
||
//3.将新链表拎出来:
|
||
//1)搞个虚拟头结点,最后返回dummy->next
|
||
//2)一有虚拟节点,那必须有一个cur节点
|
||
//3)画图:
|
||
// 1.cur->next = p->next;
|
||
// 2.cur = cur->next;
|
||
// 3.难点:需要把新旧链表彻底分开p->next = p->next->next;
|
||
// p = p->next;
|
||
/**
|
||
* Definition for singly-linked list with a random pointer.
|
||
* struct ListNode {
|
||
* int val;
|
||
* ListNode *next, *random;
|
||
* ListNode(int x) : val(x), next(NULL), random(NULL) {}
|
||
* };
|
||
*/
|
||
class Solution {
|
||
public:
|
||
ListNode *copyRandomList(ListNode *head) {
|
||
|
||
}
|
||
};
|
||
|
||
```
|
||
|
||
|
||
|
||
|
||
|
||
## 22.53. 最小的k个数
|
||
|
||
|
||
|
||
```cpp
|
||
//使用大根堆priority_queue
|
||
//只放k个,多了就踢了
|
||
//最后记翻转
|
||
class Solution {
|
||
public:
|
||
vector<int> getLeastNumbers_Solution(vector<int> input, int k) {
|
||
priority_queue<int>heap;
|
||
for(auto x : input)
|
||
{
|
||
heap.push(x);
|
||
if(heap.size() > k)heap.pop();//把堆顶删了
|
||
}
|
||
vector<int>res;
|
||
while(heap.size())
|
||
{
|
||
res.push_back(heap.top());
|
||
heap.pop();
|
||
}
|
||
reverse(res.begin(),res.end());
|
||
return res;
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 23.33. 链表中倒数第k个节点
|
||
|
||
```cpp
|
||
//先求链表长度n
|
||
//链表中倒数第k个节点 == 链表从前往后挪n-k次
|
||
//简单画个图
|
||
/**
|
||
* Definition for singly-linked list.
|
||
* struct ListNode {
|
||
* int val;
|
||
* ListNode *next;
|
||
* ListNode(int x) : val(x), next(NULL) {}
|
||
* };
|
||
*/
|
||
class Solution {
|
||
public:
|
||
ListNode* findKthToTail(ListNode* pListHead, int k) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 24.71. 二叉树的深度
|
||
|
||
```cpp
|
||
//max(左子树,右子树)+1
|
||
//dfs
|
||
//一共两行代码
|
||
/**
|
||
* Definition for a binary tree node.
|
||
* struct TreeNode {
|
||
* int val;
|
||
* TreeNode *left;
|
||
* TreeNode *right;
|
||
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
|
||
* };
|
||
*/
|
||
class Solution {
|
||
public:
|
||
int treeDepth(TreeNode* root) {
|
||
if(!root)return 0;
|
||
return max(treeDepth(root->left),treeDepth(root->right))+1;
|
||
}
|
||
};
|
||
```
|
||
|
||
## 25.72. 平衡二叉树
|
||
|
||
```cpp
|
||
//这个题和求树的最大深度一样
|
||
/*
|
||
if(!root)return 0;
|
||
*/
|
||
/**
|
||
* Definition for a binary tree node.
|
||
* struct TreeNode {
|
||
* int val;
|
||
* TreeNode *left;
|
||
* TreeNode *right;
|
||
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
|
||
* };
|
||
*/
|
||
class Solution {
|
||
public:
|
||
bool isBalanced(TreeNode* root) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
```cpp
|
||
//错误的做法:只能保证根节点的左右子树,相差不超过1,不能拿保证"任意结点的左右子树的深度相差不超过 1"
|
||
// //树的深度
|
||
// class Solution {
|
||
// public:
|
||
// int dfs(TreeNode* root)
|
||
// {
|
||
// if(!root)return 0;
|
||
// return max(dfs(root->left),dfs(root->right))+1;
|
||
// }
|
||
// bool isBalanced(TreeNode* root) {
|
||
// if(!root)return true;
|
||
// int l = dfs(root->left);
|
||
// int r = dfs(root->right);
|
||
// //cout<< l<<' '<<r<<' ';
|
||
// if(abs(l-r)>1)return false;
|
||
// return true;
|
||
// }
|
||
// };
|
||
|
||
|
||
|
||
//正确的:他计算了任意一个节点的左右子树的高度不超过1
|
||
class Solution {
|
||
public:
|
||
bool res = true;
|
||
int dfs(TreeNode* root)
|
||
{
|
||
if(!root)return 0;
|
||
int left = dfs(root->left) ,right = dfs(root->right);
|
||
if(abs(left-right)>1)res = false;
|
||
return max(left,right)+1;
|
||
}
|
||
bool isBalanced(TreeNode* root) {
|
||
dfs(root);
|
||
return res;
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 26.15. 二维数组中的查找
|
||
|
||
```cpp
|
||
//每列是递增的
|
||
//难点:右上角
|
||
//(i,j)=>(0,array[0].size()-1)
|
||
//在范围内:(i<array.size() && j>=0)
|
||
//x = array[i][j],直接返回true
|
||
//x > target j--;
|
||
//else i++;
|
||
class Solution {
|
||
public:
|
||
bool searchArray(vector<vector<int>> array, int target) {
|
||
|
||
}
|
||
};
|
||
```
|
||
|
||
## 27.59. 把数字翻译成字符串
|
||
|
||
```cpp
|
||
//dp的问题:直接记下来就行了
|
||
class Solution {
|
||
public:
|
||
int getTranslationCount(string s) {
|
||
int n = s.size();
|
||
vector<int> f(n + 1);//因为dp从1开始
|
||
f[0] = 1;
|
||
for (int i = 1; i <= n; i ++ ) {
|
||
f[i] = f[i - 1];//f[i]包含了f[i - 1]的所有情况
|
||
if (i > 1) {
|
||
int t = s[i - 1] - '0' + (s[i - 2] - '0') * 10;//状态转移的条件:只有[10,25]才可以状态转移
|
||
if (t >= 10 && t <= 25) f[i] += f[i - 2];//条件all满足:f[i] = f[i - 1]+f[i - 2];
|
||
}
|
||
}
|
||
return f[n];
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
##
|
||
|
||
## 29.37. 树的子结构
|
||
|
||
```cpp
|
||
以根为开始,开始进行p1与p2的匹配:
|
||
需要另写一个函数进行匹配,函数的参数是以p1,p2为根节点的一段小树
|
||
函数内部实现:
|
||
1.p2为空.说明之前的匹配好了,直接return true;
|
||
2.p1为空 或者 p1和p2的值不匹配,直接return false;
|
||
3.没有被条件1和2直接return,说明当前的节点是匹配的,我们接着
|
||
匹配左左和右右,是且的关系
|
||
class Solution {
|
||
public:
|
||
bool dfs(TreeNode* p1, TreeNode* p2)
|
||
{
|
||
if(!p2)return true;
|
||
if(!p1 || p1->val != p2->val)return false;
|
||
//此时,当前节点是匹配的
|
||
return dfs(p1->left,p2->left) && dfs(p1->right,p2->right);//左边匹配且右边匹配
|
||
}
|
||
bool hasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {
|
||
if(!pRoot1 || !pRoot2)return false;
|
||
if(dfs(pRoot1,pRoot2))return true;//遍历当前节点
|
||
return hasSubtree(pRoot1->left,pRoot2) || hasSubtree(pRoot1->right,pRoot2);//下一个节点
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 30.46. 二叉搜索树的后序遍历序列
|
||
|
||
```cpp
|
||
//和那个重建二叉树思路有一点点类似
|
||
//画图确定范围
|
||
class Solution {
|
||
public:
|
||
//1.搞一个全局变量
|
||
vector<int> seq;
|
||
|
||
bool verifySequenceOfBST(vector<int> sequence) {
|
||
seq = sequence;
|
||
return dfs(0, seq.size() - 1);//dfs的是范围
|
||
}
|
||
|
||
//dfs的是范围
|
||
bool dfs(int l, int r) {
|
||
if (l >= r) return true; // 1.如果子序列为空或只有一个节点，它必然是BST的后序遍历结果
|
||
int root = seq[r]; // 2.根节点 子序列的最后一个节点是当前子树的根节点
|
||
int k = l;//左右子树的分界线(找到右子树的第一个元素)
|
||
while (k < r && seq[k] < root) k++; // 找到左子树和右子树的分界点
|
||
for (int i = k; i < r; i++)//遍历右子树,如果右子树中,有元素小于根节点就不满足二叉搜索树
|
||
if (seq[i] < root)
|
||
return false; // 如果在右子树中找到比根节点小的值，则不是合法的BST后序遍历
|
||
return dfs(l, k - 1) && dfs(k, r - 1); // 递归检查左子树和右子树
|
||
}
|
||
};
|
||
|
||
```
|
||
|
||
|
||
|
||
## 31.26. 二进制中1的个数
|
||
|
||
```cpp
|
||
//记住就好, x与-x的与
|
||
int lowbit(int x){
|
||
return x&(-x);
|
||
}//可以得到最后一个1
|
||
```
|
||
|
||
|
||
|
||
## 32. 49. 二叉搜索树与双向链表
|
||
|
||
|
||
|
||
|
||
|
||
## 34.32. 调整数组顺序使奇数位于偶数前面
|
||
|
||
```cpp
|
||
维护两个指针,一个从头一个从尾
|
||
使得:前指针前面都是奇数,后指针都是偶数
|
||
|
||
前指针遇到偶数就停下来,遇到奇数就前进
|
||
后指针遇到奇数就停下来,遇到偶数就前进
|
||
|
||
如何前指针的下标和后指针的下标不同,就交换两个数
|
||
while(i < j)
|
||
{
|
||
while()
|
||
while()
|
||
}
|
||
class Solution {
|
||
public:
|
||
void reOrderArray(vector<int> &array) {
|
||
int i = 0 ,j =array.size()-1;
|
||
while(i < j)
|
||
{
|
||
while(i<j && array[i] % 2 == 1)i++;
|
||
while(i<j && array[j] % 2 == 0)j--;
|
||
if(i<j)swap(array[i],array[j]);
|
||
}
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 35.60. 礼物的最大价值
|
||
|
||
|
||
|
||
## 36.63. 字符串中第一个只出现一次的字符
|
||
|
||
|
||
|
||
## 37.85. 不用加减乘除做加法
|
||
|
||
```cpp
|
||
class Solution {
|
||
public:
|
||
int add(int num1, int num2){
|
||
while(num2)
|
||
{
|
||
int sum = num1 ^ num2;//不进位
|
||
int carray = (num1 & num2)<<1;//进位
|
||
num1 = sum , num2 = carray;//把进位拿去加,直到加为0
|
||
}
|
||
return num1;//num1是答案
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 38.47. 二叉树中和为某一值的路径
|
||
## 39.82. 圆圈中最后剩下的数字
|
||
|
||
```cpp
|
||
编号: 旧 = (新+m)%n
|
||
f(n,m) = (f(n-1,m)+m)%n;
|
||
class Solution {
|
||
public:
|
||
int lastRemaining(int n, int m){
|
||
if(n == 1)return 0;
|
||
return (lastRemaining(n-1,m)+m)%n;
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
|
||
## 40.39. 对称的二叉树
|
||
## 41.58. 把数组排成最小的数
|
||
## 42.38. 二叉树的镜像
|
||
##
|
||
|
||
|
||
## 44.43. 不分行从上往下打印二叉树
|
||
## 45.45. 之字形打印二叉树
|
||
## 46.69. 数组中数值和下标相等的元素
|
||
##
|
||
## 48.25. 剪绳子
|
||
## 49.88. 树中两个结点的最低公共祖先
|
||
## 50.17. 从尾到头打印链表
|
||
## 51.61. 最长不含重复字符的子字符串
|
||
## 52.80. 骰子的点数
|
||
## 53.20. 用两个栈实现队列
|
||
## 54.83. 股票的最大利润
|
||
## 40. 顺时针打印矩阵
|
||
## 44. 分行从上往下打印二叉树
|
||
## 51. 数字排列
|
||
------
|
||
|
||
|
||
|
||
|
||
|
||
|