转载自：http://www.acmerblog.com/leetcode-solution-search-in-rotated-sorted-array-ii-6210.html
题目描述
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e.,0 1 2 4 5 6 7might become4 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.
题目描述
我们要编写一个类似二分搜索的程序，虽然给我们的数列是打乱的，但数列总体还是有规律的，数列类型可以分成两种，像题目中说的那样，一种是依次递增的，另一种是把前半部分放到了后面，
我找到最容易理解的代码如下，代码中测试前半部分
// LeetCode, Search in Rotated Sorted Array II
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:bool search(int A[], int n, int target) {int first = 0, last = n;while (first != last) {const int mid = (first + last) / 2;if (A[mid] == target)return true;if (A[first] < A[mid]) {if (A[first] <= target && target < A[mid])last = mid;elsefirst = mid + 1;} else if (A[first] > A[mid]) {if (A[mid] < target && target <= A[last-1])first = mid + 1;elselast = mid;} else//skip duplicate onefirst++;}return false;}
};
