第一题344.反转字符串 编写一个函数,其作用是将输入的字符串反转过来.输入字符串以字符数组 s 的形式给出. 不要给另外的数组分配额外的空间,你必须原地修改输入数组.使用 O(1) 的额外空间解决这一问题. ψ(｀∇´)ψ 我的思路 取到字符串的中点,依次交换前后两部分的位置 package string; public class ReverseString { public static void reverseString(char[] s) { char temp; for (int…

344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) { if(s.empty()) return; ; ; while(start < end){ char tmp = s[end]; s[end] = s[start]; s[start] = tmp; start++; end--; } return; } }; 541. Reverse Strin…

Question 796. Rotate String Solution 题目大意:两个字符串匹配 思路:Brute Force Java实现: public boolean rotateString(String A, String B) { if (A.length() != B.length()) return false; if (A.length() == 0) return true; // Brute Force for (int i=0; i<A.length(); i++) {…

problem 796. Rotate String solution1: class Solution { public: bool rotateString(string A, string B) { if(A.size()!=B.size()) return false; && B.size()==) return true;//errr... ; i<A.size(); ++i) { , i) == B) return true; } return false; } }; s…

一.匿名函数 以后面试或者工作中经常用匿名函数 lambda,也叫一句话函数. 课上练习: # 正常函数: def func(a, b): return a + b print(func(4, 6)) # 10 # 匿名函数: func = lambda a, b: a + b print(func(11, 33)) # 44 # 写匿名函数:接收一个可切片的数据,返回索引为0与2的对应的元素(元组形式). func = lambda x: (x[0], x[2]) print(func('Dy…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/rotate-string/description/ 题目描述 We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost charac…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

［抄题］: We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can…

class Solution { public: bool rotateString(string A, string B) { if(A.length()==B.length()&&(A+A).find(B)!=string::npos) return true; return false; } }; 把俩A拼起来,能找到B,则可以通过循环右移将A转化为B…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…