Search in Rotated Sorted Array

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：这道题算是二分查找的升级版。每次从中间阶段，至少一段有序，一段无序，如果target是在有序一段，好办，直接二分。如果target在无序一段，递归。代码如下：

class Solution {public:    int search(int A[], int n, int target) {        int L=0, R=n-1, M;while (L<=R) {M = (L+R)/2;if (target==A[M]) {return M;}if (A[L]<=A[M]) {  //[L,M]有序if (A[L]<=target && target<A[M]) {R = M - 1;}else {L = M + 1;}}else {  //[M,R]有序if (A[M]<target && target<=A[R]) {L = M + 1;}else {R = M - 1;}}}return -1;    }};