LeetCode（28）Implement strStr()

来源：互联网发布：删除数组中指定元素 c# 编辑：程序博客网时间：2024/05/27 14:13

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 had been updated to return the index instead of the pointer. If you still see your function signature returns a char * or String, please click the reload button to reset your code definition.

分析

这是一道模式匹配算法，给定两个字符串haystack与needle，给出needle在haystack完全匹配的首位置坐标（从0开始）。
这道题在数据结构中有讲解，除了开始的简单模式匹配算法BF算法，还给出了改进后的KMP经典算法，下面分别用这两个算法实现。

AC代码

class Solution {public:    //简单模式匹配算法（BF算法）    int strStr(string haystack, string needle) {        int len = strlen(haystack.c_str()), nl = strlen(needle.c_str());        int i = 0, j = 0;        while (i < len && j < nl)        {            if (haystack[i] == needle[j])            {                i++;                j++;            }            else{                i = i - j + 1;                j = 0;            }//else        }//while        if (j == nl)            return i - nl;        else            return -1;    }};

KMP算法实现

class Solution {public:    //简单模式匹配算法（BF算法）    int strStr(string haystack, string needle) {        int len = strlen(haystack.c_str()), nl = strlen(needle.c_str());        int i = 0, j = 0;        while (i < len && j < nl)        {            if (haystack[i] == needle[j])            {                i++;                j++;            }            else{                i = i - j + 1;                j = 0;            }//else        }//while        if (j == nl)            return i - nl;        else            return -1;    }    //从字符串haystack的第pos个位置开始匹配    int KMP(const string &haystack, const string &needle, int pos)    {        int len = strlen(haystack.c_str()), nl = strlen(needle.c_str());        int i = pos, j = 0;        int *n = Next(needle);        while (i < len && j < nl)        {            if (j == -1 || haystack[i] == needle[j])            {                i++;                 j++;            }            else{                j = n[j];            }//else             }//while        if (j == nl)            return i - nl;        else            return -1;    }    int* Next(const string &s)    {        int i = 0, j = -1;        int next[500] ;        int len = strlen(s.c_str());        next[0] = -1;;        while (i < len)        {            while (j >-1 && s[i] != s[j])                j = next[j];            i++;            j++;            if (s[i] == s[j])                next[i] = next[j];            else                next[i] = j;                        }//while        return next;    }};

GitHub测试程序源码

0 0