Suppose an array sorted in ascending order 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.

题目大意：

假设按照升序排列的数组在事先未知的某个关键点上旋转。（即，0 1 2 4 5 6 7可能变成4 5 6 7 0 1 2）。给你一个目标值来搜索。 如果在数组中发现它返回其索引，否则返回-1。
你可能会认为数组中没有重复。

算法分析：

给定数据可能被关键点分为俩部分连续的子数组，我们根据二分法找到中位数，判断目标数落在哪儿个子组数组内，然后在子数组内用二分法找到目标数所在数据的位置，如未找到，则返回-1.

代码如下：

• Java

public int search(int[] nums, int target) {
        if (nums.length <= 0)
            return -1;
        int start = 0;
        int end = nums.length-1;
        while (start+1 < end) {
            int mid = start + (end-start)/2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] > nums[start]) {
                if (target >= nums[start] && target<=nums[mid])
                    end = mid;
                else
                    start = mid;
            } else {
                if (target >= nums[mid] && target <= nums[end])
                    start = mid;
                else
                    end = mid;
            }
        }
        if (nums[start] == target)
            return start;
        if (nums[end] == target)
            return end;
        return -1;
    }

• Python

def search(self, nums, target):
            """
            :type nums: List[int]
            :type target: int
            :rtype: int
            """
            if not nums:
                return -1
            start, end = 0, len(nums)-1
            while start+1 < end:
                mid = start  +(int) ((end - start) / 2)
                if nums[mid] == target:
                    return mid
                elif nums[mid] > nums[start]:
                    if target>=nums[mid] and target<=nums[end]:
                        start == mid
                    else:
                        end == mid
                else:
                    if target>=nums[start] && target<=nums[mid]:
                        end == mid
                    else:
                        start == mid
            if nums[start] == target:
                return start
            if nums[end] == target:
                return end
            return -1

注： Python代码未在LeetCode验证通过，原因时间超出限制，有待优化。