[LeetCode]44. Implement strStr()实现strStr()函数

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.

解法1：暴力破解法，两重循环进行搜索，时间复杂度O(mn)。

class Solution {
public:
    int strStr(string haystack, string needle) {
        int n = haystack.size(), m = needle.size();
        if(n == 0 || m == 0)
        {
            if(n >= 0 && m == 0) return 0;
            return -1;
        }
        for(int i = 0; i < n - m + 1; i++)
        {
            int j = 0;
            for(; j < m; j++)
            {
                if(haystack[i + j] != needle[j]) break;
            }
            if(j == m) return i;
        }
        return -1;
    }
};

解法2：在haystack里面匹配needle，使用Sunday算法进行模式匹配。时间复杂度O(m+n)。

class Solution {
public:
    int strStr(string haystack, string needle) {
        int pos = -1;
        bool res = sunday(haystack, needle, pos);
        return pos;
    }
private:
    bool sunday(const string& text, const string& patt, int& pos)
    {
        int ts = text.size();
        int ps = patt.size();
        if (ts <= 0 || ps <= 0)
        {
            if (ts >= 0 && ps == 0)
            {
                pos = 0;
                return true;
            }
            return false;
        }
        vector<int> dis_tab(256, ps + 1);
        for (int i = 0; i < ps; ++i)
            dis_tab[(unsigned int)patt[i]] = ps - i;

        int end = ts - ps + 1;
        for (int i = 0; i < end;)
        {
            int j = 0;
            if (text[i] == patt[0])
                while (i < ts && j < ps && text[i + j] == patt[j]) ++j;
            if (j == ps)
            {
                pos = i + j - ps;
                return true;
            }
            i += dis_tab[text[i + ps]];
        }
        return false;
    }
};