Github leetcode 我的解题仓库   https://github.com/interviewcoder/leetcode 题目: Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Update (2014-11-02):The signature of the function ha…

LeetCode解题之Implement strStr() 原题 实现字符串子串匹配函数strStr(). 假设字符串A是字符串B的子串.则返回A在B中首次出现的地址.否则返回-1. 注意点: - 空字符串是全部字符串的子串,返回0 样例: 输入: haystack = "abc", needle = "bc" 输出: 1 输入: haystack = "abc", needle = "gd" 输出: -1 解题思路 字符串匹…

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Update (2014-11-02): The signature of the function had been updated to return the index instead of the pointer. If you still…

Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. I wrote that function before in practice without knowing anything about strStr(), so I have a pretty clear brute-force sol…

[题目] Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. [题意] 实现库函数strStr(), 功能是在字符串haystack中找出目标串needle第一次出现的索引位 [思路] 字符串的匹配,能够用暴力解法,但不推荐.一般使用KMP算法求解. 简要介绍一下KMP的思想: haystack…

Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. 这个题考查的是KMP算法.先求特征向量,然后再进行匹配,确实能够大大提高效率.code例如以下: class Solution { public: char *strStr(char *haystack, char *needle) { if(…

Implement strStr(). Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Example 1: Input: haystack = "hello", needle = "ll" Output: 2 Example 2: Input: haystack = "aaaaa",…

题目链接:https://leetcode.com/problems/implement-strstr/description/ 题目大意:字符串匹配,从字符串中,找到给定字符串第一次出现的位置下标,并返回. 法一:暴力,两个for循环,逐一比较每一个可能的字符串,一旦找到,则返回.代码如下(耗时508ms->10ms): public int strStr(String haystack, String needle) { int res = -1, len_h = haystack.leng…

#kmp class Solution(object): def strStr(self, haystack, needle): """ :type haystack: str :type needle: str :rtype: int """ if len(haystack) <= 0 and len(needle)<=0: return 0 arrNext=self.getNext(needle) i=0 j=0 intHLen=…

如题 思路:暴力就行了.1ms的暴力!!!别的牛人写出来的,我学而抄之~ int strStr(char* haystack, char* needle) { ; ; ; ++i) { ; ; ++j) { ) return i; ) ; if (haystack[i + j] != needle[j]) break; } } } strStr()…