Search in Rotated Sorted Array

题目：

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

分析：

这题目用个分类讨论的思想可以做，分为mid > end 和 mid < end两种情况。

这题目与常见的Binary search架构略有不同！！！

解法：

class Solution {
    /** 
     * param A : an integer ratated sorted array
     * param target :  an integer to be searched
     * return : an integer
     */
public:
    int search(vector<int> &A, int target) {
        // write your code here
        if (A.size() == 0) {
            return -1;
        }

        int start = 0;
        int end = A.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;

            if (A[mid] == target) {
                return mid;
            } 

            if (A[mid] < A[end]) {
                if (A[mid] < target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            } else {
                if (A[start] <= target && target < A[mid]) {
                    end = mid;
                } else {
                    start = mid;
                }
            }
        }

        if (A[start] == target) {
            return start;
        }
        if (A[end] == target) {
            return end;
        }
        return -1;
    }
};