题目：

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6]might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example1：
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example2：
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

思路：

首先需要找到数组中两个递增的序列，以 [2,5,6,0,0,1,2]为例。首先要找到6的位置这样就相当于把数组分为[2,5,6]和[0,0,1,2]两部分呢。

之后对两部分分别进行二叉查找。返回结果。

代码：

public boolean search(int[] nums, int target) {
        boolean flag = false;
        if(nums == null || nums.length == 0)
            return flag;
        int pivot = 0;
        for(int i=0;i<nums.length - 1;i++){
            if(nums[i] > nums[i+1]){
                pivot = i;
                break;
            }
        }
        return binarySearch(nums, 0, pivot, target) || binarySearch(nums, pivot + 1, nums.length - 1, target);
}

/**
     * Binary search to find target.
     * @param nums
     * @param left
     * @param right
     * @param target
     * @return
     */
private boolean binarySearch(int [] nums, int left, int right, int target){
    int begin = left;
    int end = right;
    while(begin <= end){
        int mid = (begin + end)/2;
        if(nums[mid] == target)
            return true;
        if(nums[mid] < target)
            begin = mid + 1;
        else
            end = mid - 1;
    }
    return false;
}