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).

Find the minimum element.

Notice

The array may contain duplicates.

Given [4,4,5,6,7,0,1,2] return 0.

和I類似，OOXX二分法(find the first X)，只是對比對象變掉。從原來的最後一個數，變為最後開始數起第一個比不等於nums[0]的數。

public class Solution {
    /*
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {

        //返回int這裏怎麽處理？
        if (nums == null || nums.length == 0){
            throw new IllegalArgumentException();
        }

        
 
int start = 0;
        int end = nums.length - 1;

        while (end > 0 && nums[end] == nums[start]){
            end--;
        }

        int cmpTarget = nums[end];

        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (nums[mid] > cmpTarget){
                start 
 
= mid;
            } else {
                end = mid;
            }
        }

        return Math.min(nums[start], nums[end]);
    }
}

lintcode160- Find Minimum in Rotated Sorted Array II- medium