一個系列三道題，我都不會做，google之答案。過了兩道，第三道看不懂，放置，稍后繼續。 一、Best Time to Buy and Sell Stock I 題目：一個數組表示一支股票的價格變換。要求只買賣一次，獲得最大收益。 思路：一開始我認為

---恢復內容開始--- The problem: Given two words (start?and?end), and a dictionary, find the length of shortest transformation sequence from?start?to?end, such that: Only one letter can be changed at a timeEach intermediate word must exist in the dictionar

Given an array of integers, every element appears?twice?except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? leetcode121，? 　　看到這個題目的時候，由于

題目連接 https://leetcode.com/problems/valid-palindrome/? Valid Palindrome Description Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example,?“A man, a plan, a canal: Panama” is a palin

題目描述: Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the?Hamming weight). leetcode？For example, the 32-bit integer ’11' has binary representation?00000000000000000000000000001011, s

題目鏈接: https://oj.leetcode.com/problems/add-two-numbers/ ? 問題： You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contains a single digit. Add the two numbers an

Follow up for “Unique Paths”: Now consider if some obstacles are added to the grids. How many unique paths would there be? leetcode cn。An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the

Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = “anagram”, t = “nagaram”, return true. s = “rat”, t = “car”, return false. Note: You may assume the string contains only lowercase

?7. Reverse Integer Problem's Link ?---------------------------------------------------------------------------- Mean:? 將一個整數的數值位反轉. analyse: LEETCODE，題目沒說當精度溢出時返回0.這個地方要注意一下. Time complexity: O(N) ? view code /** * --

題目： 方法一： 首先我們可以想到暴力解法，就是 逐個生成字符串 看他受否含有重復字符 如下代碼暴力法： int LenOfUniqueStr(char* Start) {int Validity[128] = {0};int Len = 0;char* ReadPos = Start;for (; *ReadPos != '