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;
}
};