Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().

题目大意：

  给定两个字符串haystack和needle，找出needle在haystack中首次出现的位置。如果没找到则返回-1，并且当needle是空串的时候返回0。

解题思路：

  简单呐~~~~~~~~~~~~~~~~直接haystack.find(needle)（手动滑稽

  也确实是简单，要查找肯定需要一次遍历，没什么好说的。

  从头开始遍历，每次遍历取临时遍历temp与当前遍历下标相同，若遍历needle.size()次后完全匹配则直接返回当前的遍历下标即可

解题代码：

class Solution {
public:
    int strStr(string haystack, string needle) {
    //空串返回0
    if (needle.empty())
        return 0;
    //needle长度比haystack大时返回-1
    if (needle.size() > haystack.size())
        return -1;
    size_t haySize = haystack.size(), needSize = needle.size();
    int res = -1, temp;
    //循环不变量是剩余未遍历字符长度与needle长度之和小于或等于haystack长度
    for (size_t hayIndex = 0, needIndex = 0; hayIndex + needSize <= haySize; ++hayIndex, needIndex = 0)
    {
        temp = hayIndex;
        //注意needIndex得小于needSize，不然会越界
        while (needIndex < needSize && haystack[temp] == needle[needIndex]) {
            ++temp;
            ++needIndex;
        }
        if (needIndex == needSize)
        {
            res = hayIndex;
            break;
        }
    }
    return res;
  }
};