lintcode 字符串查找

2016-12-13  本文已影响85人  yzawyx0220

对于一个给定的 source 字符串和一个 target 字符串,你应该在 source 字符串中找出 target 字符串出现的第一个位置(从0开始)。如果不存在,则返回 -1。

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class Solution {
public:
    /**
     * Returns a index to the first occurrence of target in source,
     * or -1  if target is not part of source.
     * @param source string to be scanned.
     * @param target string containing the sequence of characters to match.
     */
    int strStr(const char *source, const char *target) {
        // write your code here
        int result = -1;  
        int i = 0;  
        int j = -1;  
        if (source == nullptr || target == nullptr) {  
            return result;  
        }  
        if ( *target == '\0')  
            return i;  
        while (*source !='\0') {  
            ++j;  
            if(*source == *target) {  
                i = j;  
                const char *p = source;  
                const char *q = target;  
            while (*p == *q) {  
                if (*(q+1) =='\0') {  
                    return i;  
                }  
                p++;  
                q++;  
                }  
            }  
            ++source;  
                ++i;  
        }  
        return result;  
    }
};
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