leetcode28. Implement strStr()

最新推荐文章于 2021-10-20 10:58:48 发布

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本文介绍两种实现strStr()的方法：一种是简单的暴力匹配，时间复杂度为O(M*N)；另一种是利用KMP算法进行快速匹配。文中详细展示了两种方法的C++代码实现，并解释了KMP算法中next数组的构建过程。

easy程度题

题目：

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

字符串匹配，暴力匹配复杂度O(M * N)

AC解:

class Solution {
public:
    int strStr(string haystack, string needle) 
    {//暴力匹配
        int lengthA = haystack.length();
        int lengthB = needle.length();
        if (lengthB == 0)
            return 0;
        if (lengthA < lengthB)
            return -1;
        for (int i = 0; i <= lengthA - lengthB; i++)
        {
            int count = 0;
            while (count < lengthB && haystack[i + count] == needle[count])
                count++;
            if (count == lengthB)
                return i;
        }
        
        return -1;
    }
};

KMP快速匹配

class Solution {
public:
    int strStr(string haystack, string needle) 
    {//KMP快速匹配算法
        int lengthA = haystack.length();
        int lengthB = needle.length();
        if (lengthB == 0)
            return 0;
        if (lengthA < lengthB)
            return -1;
        int *next = new int [lengthB];
        
        makeNext(needle,next);
        return KMPSearch(haystack,needle,next);
    }
    
    inline void makeNext(string p,int *next)
    {
        int len = p.length();
        next[0] = -1;
        int k = -1, j = 0;
        while (j < len - 1)
        {
            while (k >= 0 && p[j] != p[k])
                k = next[k];
            j++;
            k++;
            if (p[j] == p[k])
                next[j] == next[k];
            else
                next[j] = k;
        }
    }
    
    inline int KMPSearch(string s, string p, int *next)
    {
        int i = 0, j = 0;
        int slen = s.length(), plen = p.length();
        while (i < slen && j < plen)
        {
            if (j == -1 || s[i] == p[j])
            {
                i++;
                j++;
            }
            else
                j = next[j];
        }
        
        if (j == plen)
            return i - j;
        else
            return -1;
    }
};

具体思想参见其它博文，刚学的时候花了几个小时弄懂了，过段时间再次遇到还是写不出代码来- -