[LintCode] 字符串问题

1、字符串包含问题

1）串的模式匹配算法

（LintCode：字符串查找）

BF

class Solution {
public:
    /*
     * @param source: source string to be scanned.
     * @param target: target string containing the sequence of characters to match
     * @return: a index to the first occurrence of target in source, or -1  if target is not part of source.
     */
    int strStr(const char *source, const char *target) {
        // write your code here
        if (source == NULL || target == NULL)
            return -1;
        int slen = strlen(source);
        int tlen = strlen(target);
        if (tlen == 0)
            return 0;
        if (slen == 0)
            return -1;
        for (int i = 0; i < slen; ++i)
        {
            for (int j = 0, s = i; j < tlen && s < slen; ++j, ++s)
            {
                if (source[s] != target[j])
                {
                    break;
                }
                else
                {
                    if (j == tlen - 1)
                        return i;
                }
                
            }
        }
        return -1;
    }
};

KMP

class Solution {
public:
    /*
     * @param source: source string to be scanned.
     * @param target: target string containing the sequence of characters to match
     * @return: a index to the first occurrence of target in source, or -1  if target is not part of source.
     */
    int strStr(const char *source, const char *target) {
        // write your code here
        if (source == NULL || target == NULL)
            return -1;
        int slen = strlen(source);
        int tlen = strlen(target);
        if (tlen == 0)
            return 0;
        if (slen == 0)
            return -1;
        return kmp(source, target);
    }
    
    int kmp(const char *source, const char *target)
    {
        int *next = getNext(target);
        int slen = strlen(source);
        int tlen = strlen(target);
        int i = 0, j = 0, res = -1;
        while (i < slen)
        {
            if (j == -1 || source[i] == target[j])
            {
                ++i;
                ++j;
            }
            else
            {
                j = next[j];
            }
            if (j == tlen)
            {
                res = i - tlen;
                break;
            }
        }
        delete[] next;
        return res;
    }
    
    int* getNext(const char *target)
    {
        int tlen = strlen(target);
        int *next = new int[tlen];
        int i = 0, j = -1;
        next[0] = -1;
        while (i < tlen - 1)
        {
            if (j == -1 || target[i] == target[j])
            {
                ++i;
                ++j;
                next[i] = j;
            }
            else
            {
                j = next[j];
            }
        }
        return next;
    }
};