Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times. [0,1,2,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], …, a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], …, a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

<解題>

1. 看到sorted array，可以想到用Binary search解題
2. if( midpoint > 0 && nums[midpoint] < nums[midpoint-1] )，因為當前那個點的值<前面index的點的值，代表找到那個點，因此return nums[midpoint] 3.如果：

• 左側都是sorted&&右側沒有sorted：找右側
• else：找左側

class Solution {
    public int findMin(int[] nums) {
        if(nums.length ==0) return -1;
        if(nums.length ==1) return nums[0];

        int left = 0;
        int right = nums.length -1;

        while(left<right){
            int midpoint = left + (right - left)/2;
            if( midpoint > 0 && nums[midpoint] < nums[midpoint-1] ){
                return nums[midpoint];
            } else if (nums[left]<= nums[midpoint] && nums[midpoint] > nums[right] ){ //sorted on the left side && !=sorted on the right side
                left = midpoint +1 ;
            } else {
                right = midpoint -1;
            }
        }
        return nums[left];
    }
}

Time: O(log n)

Space: O(1)