[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.

判读一个字符串是否是另一个字符串的子串~可以用brute force的方法来解

Leetcode提供的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:
    # @param haystack, a string
    # @param needle, a string
    # @return an integer
    def strStr(self, haystack, needle):
        for i in xrange(len(haystack) - len(needle) + 1):
            existFlag = True
            for j in xrange(len(needle)):
                if haystack[i + j] != needle[j]:
                    existFlag = False
                    break
            if existFlag == True:
                return i
        return -1