28. Implement strStr()

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.

class Solution {
public:
int strStr(string haystack, string needle) {
int hLen = haystack.length();
int nLen = needle.length();

int i = 0;
int j = 0;
while(i<hLen && j<nLen)
{
if(haystack[i] == needle[j])
{
//如果当前字符匹配成功（即S[i] == P[j]），则i++，j++
++i;
++j;
}
else
{
//如果失配（即S[i]! = P[j]），令i = i - (j - 1)，j = 0
//这步是关键,i指向匹配失败的下一个元素
i = i-j+1;
j = 0;
}
}
//匹配成功，返回模式串p在文本串s中的位置，否则返回-1
if(j == nLen)
return i-j;
else
return -1;

}
};

class Solution {
public:
int strStr(char *haystack, char *needle) {

int sLen = strlen(haystack);
int pLen = strlen(needle);

int i = 0;
int j = 0;
while (i < sLen && j < pLen)
{
if (haystack[i] == needle[j])
{
//①如果当前字符匹配成功（即S[i] == P[j]），则i++，j++
i++;
j++;
}
else
{
//②如果失配（即S[i]! = P[j]），令i = i - (j - 1)，j = 0
//这步是关键
i = i - j + 1;
j = 0;
}
}
//匹配成功，返回模式串p在文本串s中的位置，否则返回-1
if (j == pLen)
return i - j;
else
return -1;

}

};