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.

思路： 不同于普通的B search， 此题需要判断左右哪一部分是正常没有折叠过的 （有且只有一边没有折叠）。

代码： 

class Solution {
public:
    int search(int A[], int N, int key) {
    int L = 0;
    int R = N - 1;

    while (L <= R) {
        // Avoid overflow, same as M=(L+R)/2
        int M = L + ((R - L) / 2);
        if (A[M] == key) return M;

        // the bottom half is sorted
        if (A[L] <= A[M]) {
            if (A[L] <= key && key < A[M])
                R = M - 1;
            else
                L = M + 1;
        }
        // the upper half is sorted
        else {
            if (A[M] < key && key <= A[R])
                L = M + 1;
            else 
                R = M - 1;
        }
    }
    return -1;
}
};