bettle_king的博客

https://leetcode-cn.com/problems/maximum-depth-of-n-ary-tree//*// Definition for a Node.class Node {public: int val; vector<Node*> children; Node() {} Node(int _val) { val = _val; } Node(int _val, vector<Nod.

https://leetcode-cn.com/problems/delete-node-in-a-linked-list/要想自己删自己，只需替换自己的值为儿子的值，删除儿子即可。/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */cla.

题目地址class Solution {public: void nextPermutation(vector<int>& nums) { next_permutation(nums.begin(),nums.end()); }};正规解法:

题目地址class Solution {public: int divide(int dividend, int divisor) { long long int temp=1ll*dividend/divisor; if(temp>INT_MAX) return INT_MAX; return dividend/divisor; }};正解：用二进制倍增的思想也就是说x可以减k次y。我们将k弄成二进制，预处理每一个数。.

题目地址class Solution {public: int strStr(string haystack, string needle) { return haystack.find(needle); }};KMP方法：class Solution {public: int strStr(string s, string p) { if(p.empty()) return 0; int n=s.size(),m=p..

题目地址class Solution {public: int removeElement(vector<int>& nums, int val) { int k=0; for(int i=0;i<nums.size();i++) if(nums[i]!=val) nums[k++]=nums[i]; return k; }};

题目地址class Solution {public: int removeDuplicates(vector<int>& nums) { int ans=0; int t=999999; for(int i=0;i<nums.size();i++) { if(t==nums[i]) continue; else { .

题目地址/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next.

题目地址我这种方法不是正解，因为交换的只是值，真正的指针并没有交换。/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x,.

题目地址/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next.

题目地址class Solution {public: vector<string> ve; vector<string> generateParenthesis(int n) { dfs(1,n,""); return ve; } void dfs(int index,int n,string s) { if(index==n*2+1) { stack<char&gt.

题目地址/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next.

题目地址非常裸的一个关于栈的应用class Solution {public: bool isValid(string s) { stack<char> st; map<char,char> mp; mp[')']='('; mp[']']='['; mp['}']='{'; for(int i=0;i<s.size();i++) { if(st.size()==0) s.

题目地址/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next.

题目地址考察的双指针，先排序，用三层循环，里面用一个双指针。class Solution {public: vector<vector<int>> fourSum(vector<int>& nums, int target) { sort(nums.begin(),nums.end()); vector<vector<int>> ans; for(int i=0;i<nums.s.

题目地址暴力模拟：class Solution {public: vector<string> letterCombinations(string digits) { string s[15]; s[0]="0",s[1]="!@#",s[2]="abc",s[3]="def",s[4]="ghi",s[5]="jkl"; s[6]="mno",s[7]="pqrs",s[8]="tuv",s[9]="wxyz"; vector.

题目地址方法一: 双指针，如果k向后移动其差值反而变大了那么就breakclass Solution {public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(),nums.end()); int sum=99999999; int temp=0; for(int i=0;i<nums.size();i++) .

题目链接最后两点TLE做法class Solution {public: vector<vector<int>> threeSum(vector<int>& nums) { map< vector<int> , bool > mp; unordered_map<int,int>hush; vector<vector<int>> ans; .

https://leetcode-cn.com/problems/longest-common-prefix/class Solution {public: string longestCommonPrefix(vector<string>& strs) { string ans; if(strs.size()==1) return strs[0]; int k=0; while(1) { .

题目地址class Solution {public: int romanToInt(string s) { int ans=0; map<char,int> mp; mp['I']=1; mp['V']=5; mp['X']=10; mp['L']=50; mp['C']=100; mp['D']=500; mp['M']=1000; .

https://leetcode-cn.com/problems/integer-to-roman/class Solution {public: string intToRoman(int num) { string ans; int m_t=num/1000; for(int i=1;i<=m_t;i++) ans+="M"; num=num%1000; if(num>=900) ans+="CM",num-.

题目地址用双指针，暴力一定会超时。木桶容量由短板和距离决定, 移动长板的话, 水面高度不可能再上升, 而宽度变小了, 所以只有通过移动两者中的短板, 才有可能使水位上升。class Solution {public: int maxArea(vector<int>& height) { int ans=0; int i=0,j=height.size()-1; while(i<j) { .

题目跳转class Solution {public: bool isPalindrome(int x) { if(x<0) return false; string a=to_string(x); int n=a.size(); for(int i=0;i<n/2;i++) if(a[i]!=a[n-i-1]) return false; return true; }.

题目链接class Solution {public: int myAtoi(string s) { string ans; bool flag=false; while(s[0]==' '&&s.size()>1) s=s.substr(1);//去除前导零 if(s[0]=='-') flag=true,s=s.substr(1); else if(s[0]=='+') s=s.substr(1);.

https://leetcode-cn.com/problems/reverse-integer/class Solution {public: int reverse(long long int x) { bool flag=false; if(x<0) flag=true,x=-x; long long int ans=0; while(x) { ans=ans*10+x%10; .

https://leetcode-cn.com/problems/zigzag-conversion/找规律的题：本质和弹弹球一样的，到底又回来，到顶又下去。class Solution {public: string convert(string s, int numRows) { string ss[1005]; int cnt=0; int k=0; bool flag=true; if(numRows==.

https://leetcode-cn.com/problems/longest-palindromic-substring/方法一: 最暴力做法, 时间复杂度O(n3)class Solution {public: string longestPalindrome(string s) { int ans=0; string ss; for(int i=0;i<s.size();i++)//枚举左端点 { .

https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/双指针滑动窗口class Solution {public: int lengthOfLongestSubstring(string s) { int ans=0; map<char,int> mp; for(int i=0,j=0;i<s.size();i++) .

https://leetcode-cn.com/problems/add-two-numbers/本题其实很简单，就是一个高精度加法，只不过其用的是链表来存储每一位。这对于新手来说是十分的不友好的。于是我们来分析一下：题目给的是一个结构体，里面一个结点的值的变量和一个结点的下一个指针的变量。其中还有这两个值的初始化构造函数。首先如何定义一个结点呢？auto head=new ListNode(-1); 这种是用的智能指针(即根据当前的情况来自动的判断指针的类型)来定义，且ListNo.

https://leetcode-cn.com/problems/two-sum/方法一：暴力 时间复杂度O(n2)class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { for(int i=0;i<nums.size();i++) { for(int j=i+1;j<nums.size();j++.