Jason Zhou

Given an integer, convert it to a roman numeral.Input is guaranteed to be within the range from 1 to 3999.把一个数字变成罗马数字，直接递归暴力，代码有点丑23333class Solution {public: string intToRoman(int num) {

Write a function to find the longest common prefix string amongst an array of strings.求最长公共前缀，直接暴力class Solution {public: string longestCommonPrefix(vector<string>& strs) { string s = "";

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Note: The solution set must not contain duplic

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly

Given a digit string, return all possible letter combinations that the number could represent.A mapping of digit to letters (just like on the telephone buttons) is given below.Input:Digit string “23”

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.Note: The solution set mu

欢迎使用Markdown编辑器写博客本Markdown编辑器使用StackEdit修改而来，用它写博客，将会带来全新的体验哦：Markdown和扩展Markdown简洁的语法代码块高亮图片链接和图片上传LaTex数学公式UML序列图和流程图离线写博客导入导出Markdown文件丰富的快捷键快捷键加粗 Ctrl + B 斜体 Ctrl + I 引用 Ctrl

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linke

Given a linked list, remove the nth node from the end of list and return its head.For example,Given linked list: 1->2->3->4->5, and n = 2.After removing the second node from the end, the linked list be

Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.The brackets must close in the correct order, “()” and “()[]{}” are all valid but “

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists./** * Definition for singly-linked list. * struct ListNode

Given a string, find the length of the longest substring without repeating characters.Examples:Given “abcabcbb”, the answer is “abc”, which the length is 3.Given “bbbbb”, the answer is “b”, with the le

There are two sorted arrays nums1 and nums2 of size m and n respectively.Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).Example 1: nums1 = [1, 3] num

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 这两题都是给你一个字符串，求出最长回文子串。在以前

The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H N A P L S I I G

Reverse digits of an integer.Example1: x = 123, return 321 Example2: x = -123, return -321水题没啥好说的，就一个小陷阱，如果转换后的数大于int的范围就return0。class Solution {public: int reverse(int x) { if(x == 0)

Implement atoi to convert a string to an integer.Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.N

Determine whether an integer is a palindrome. Do this without extra space.没啥好说，判断一个数字是否为回文数字。class Solution {public: bool isPalindrome(int x) { int newx = 0; int xx = x; wh

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lin

这题1刷只想到用新的点，2刷试试不用新的点，直接在原链表中运行。 Ps 链表的申请 复习/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * };

一开始以为很难， 其实就是递归判断，只要第一次做错的原因是我以为分割点是中间，其实是每一个点都可以是分割点，想想2叉树就想懂的。 2刷应该就用dp吧， 这个也在手动刷一次！理解多一次class Solution {public: bool isScramble(string s1, string s2) { if(s1 == s2) return true;

案例很奇怪的一题，其实超级简单，但是奇怪，2刷看代码就ok，不用刷了class Solution {public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { int i = m - 1; int j = n - 1; int k = m + n - 1

规律题， 如果模拟的话或者递归的话应该超级烦 2刷可以看看其他方法，个人觉得不用刷class Solution {public: vector<int> grayCode(int n) { vector<int>ve; ve.push_back(0); if(n == 0) return ve; ve.push_back(1)

这题我是改了之前的那个就ac了，2刷一定要刷，跟第一道一起刷，因为这个dfs有问题的，会出现重复，需要改！！！class Solution {public: vector<vector<int>>fve; set<vector<int>>ve; set<vector<int>>ve1; vector<int>vee; void dfs(vector<int>

题意易懂，就是烦，要判断m是否是1，2刷想个更好的方法，最好是什么情况都通用的方法。/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */

无敌代码暴力。。。。。，2刷还是乖乖dfs吧class Solution {public: vector<string> restoreIpAddresses(string s) { vector<string> ret; string ans; for (int a=1; a<=3; a++) for (int b=1; b<

简单题， 中序遍历/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };

思路跟96一样，就是烦了一点，在右边的时候需要加上n，注意一点！！错了很多的 就是要new 新的点！！！别重复使用！！！每次生成都要new，2刷可以刷，应该有其他做法！/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNod

找规律，对于每一次的n，都可以看成有不同的数目作为根节点，然后再算这个根节点的左右有多少的不同的节点，就有多少种可能，最后左右相乘就okclass Solution {public: int numTrees(int n) { vector<int>ve; ve.push_back(1); ve.push_back(1); v

dfs: 题意就是s1和s2是否可以交叉变成s3，一开始我使用dfs，但是超时了 然后剪枝，加入一个标记数组，如果dfs过的就不进行dfs，然后就ac了class Solution {public: int mapp[500][500]; int mark = 0; void pipei(string s1, string s2, string s3, int a, i

这题要多想，首相二叉平衡树的判断一定要从全局来看，分别大于小于某个数！！！，第二点，我被最大最小值卡了，所以就用了long long int然后改最大最小值，2刷刷好看一点的代码，思路要更加清晰/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; *

做这题的时候发现其实上一题的做法是错误的，BST的中序排序是一个上升序列，所以判断的时候只需要按中序排序， 然后判断是不是上升的就ok！所以98也是这样做的！！！ 至于这题，为什么n1 和 n2 记录的不一样呢？ N1 是记录pre的而n2是记录root的 因为你在纸上写一个上升序列，然后换砖两个数，你就会发现原因！！！/** * Definition for a binary tree no

没什么好说的，判断两个数是不是一样，简单题/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NUL

必须看看学习一下代码，！！！！！！ 可以的这一题！！/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), rig

简单题 bfs搞定/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };

跟上一题一样，就最后换了vector的顺序。。。下次就应该在queue中换顺序吧/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), le

求树的深度，简单题/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };

必须要学习的代码，很骚！/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} *

单向bfs 方法也是想到和写了出来，就是没有用find这个方法导致超时！！！用了iterator 其实就是bfs不断判断是否是那就ok了，数据结构的使用方法还是有点弱！class Solution {public: void addnext(string s, queue<string>& qu, unordered_set<string>& wordList){ wo

有意思的一题，首先，时间需要时on，这个时间想到的只有hashmap，treemap不行，因为tree的操作的ologn，hasdmap是o1，然后遍历一次。如果这个数的左右两边没数，则·dp=1 有数的话，就更新一段联系区间的最左右的端点的dp值，dp值表示包括这个数在内的连续最大长度。所以时间上应该是on，其实我觉得是o3n，不过好像题解就是这个方法。。。。 2刷学习一下别人6巷代码的写法