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 find函数 遍历+切片 日期 题目地址:https://leetcode.com/problems/implement-strstr/description/ 题目描述 Implement strStr(). Return the index of the first occurrence of needle in haystack, or -1…

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 问题:实现 strStr() 函数.即在  haystack 中匹配 needle 字符串. 可以理解为,实际上这道题是在问如何实现 KMP(Knuth–Morris–Pratt) 算法.这是个效率比较高的算法,只需要扫一遍 haystack 就可…

28. Implement strStr() Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 思路:子串匹配,朴素匹配.…

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",…

题目描述: 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…

Brute Force算法,时间复杂度 O(mn) def strStr(haystack, needle): m = len(haystack) n = len(needle) if n == 0: return 0 if m < n: return -1 for i in range(m - n - 1): for j in range(n): if haystack[i + j] != needle[j]: break elif j == n - 1: return i return -1…

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",…

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",…

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 解法: 这道题让我们在一个字符串中找另一个字符串第一次出现的位置,那我们首先要做一些判断,如果子字符串为空,则返回0,如果子字符串长度大于母字符串长度,则返回-1.然后我们开始遍历母字符串,我们并不需要遍历整个母字符串,而是遍历到剩下的长度和子字符…

题目链接: https://leetcode.com/problems/implement-strstr/?tab=Description   Problem : 实现找子串的操作:如果没有找到则返回-1  ,如果找到则返回第一个匹配位置第一个字符的下标   遍历操作,依次遍历子串中的元素,从长串的第i个元素位置开始,当成功遍历结束子串所有元素时,返回此时的i即时所求结果, 如果此时的i+j已经等于长串的个数,那么表示没有找到,返回-1    参考代码: package leetcode_50;…