[leetcode]_Interleaving String】的更多相关文章

下午去蹭了一发新浪的笔试. 炒鸡多的网络基础知识,总共18道题,就写了8道左右吧,剩下的全是网络知识,这部分抽时间至少过一过. 其中一道算法题,回来跟嘟嘟商量,才发现是leetcode上的原题,连example都没有变,这可是道难度系数5的题,我嘞个去. 题目:给定三个字符串s1,s2,s3,判断s3是否能由s1和s2交错而成. 思路: 1.当s1当前字符 = s2当前字符 && s2当前字符 != s3当前字符 时,消s1. 2.同理状况 消s2. 3.难点在于当s1当前字符 = s2当…
LeetCode--Reverse String Question Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh". Answer class Solution { public: string reverseString(string s) { string str(s.…
Leetcode 8. String to Integer (atoi) atoi函数实现 (字符串) 题目描述 实现atoi函数,将一个字符串转化为数字 测试样例 Input: "42" Output: 42 Input: " -42" Output: -42 Input: "4193 with words" Output: 4193 Input: "words and 987" Output: 0 详细分析 这道题的cor…
有关这样的字符串的题真是层出不穷啊,并且他们都有这样一个特点,就是递归的思路如此简单,但一定超时! 这个时候,dp就朝我们缓缓走来.递归超,dp搞!这道题的状态转移方程还是比較好写的,用ispart[i][j]代表s1贡献i长,s2贡献j长时,能不能形成s3的前i+j个字符.更新能够依照行或者列開始,s3的前i+j个字符,能够是((i-1)+1)+j构成,也能够是i+((j-1)+1)构成,这取决于当前的这个字符s3[i+j-1]跟s1[i-1]和s2[j-1]是否相等,或的关系就能够解决这个问…
Given a non-empty string, encode the string such that its encoded length is the shortest. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note: k will be a positive integ…
Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…
Given a non-empty string str and an integer k, rearrange the string such that the same characters are at least distance k from each other. All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empt…
Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh". 这道题没什么难度,直接从两头往中间走,同时交换两边的字符即可,参见代码如下: 解法一: class Solution { public: string reverseString(string s) { , right =…
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example, Given: s1 = "aabcc", s2 = "dbbca", When s3 = "aadbbcbcac", return true. When s3 = "aadbbbaccc", return false. 这道求交织相错的字符串和之前…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…