LEETCODE-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.

不小心解鎖了一種新方法：利用了substr()函數從haystack中的第一個元素到（haystack.size() - needle.size()）位置去截取needle.size()長度的字元串，并以次将其與needle進行比較，如相等則傳回截取的起始位置；

**My solution

O(nm) runtime, O(1) space – Brute force:

You could demonstrate to your interviewer that this problem can be solved using known efficient algorithms such as Rabin-Karp algorithm, KMP algorithm, and the Boyer- Moore algorithm. Since these algorithms are usually studied in an advanced algorithms class, it is sufficient to solve it using the most direct method in an interview – The brute force method.

The brute force method is straightforward to implement. We scan the needle with the haystack from its first position and start matching all subsequent letters one by one. If one of the letters does not match, we start over again with the next position in the haystack.

The key is to implement the solution cleanly without dealing with each edge case separately.

**

class Solution {
public:
    int strStr(string haystack, string needle) {
        int len1 = haystack.size();
    int len2 = needle.size();
    if (len1 < len2)
        return -;
    int n = len1 - len2;
    for (int i = ; i <= n; i++) {
        string model = haystack.substr(i, len2);
        if (model == needle)
            return i;
    }
    return -;
    }
};