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