LC.154. Find Minimum in Rotated Sorted Array II

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/
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).

Find the minimum element.

You may assume no duplicate exists in the array.
注意，我们要找的最小值，只会在右端，所以如果已经在右端，则挪RIGHT
如果在左（通过和 NUMS[RIGHT] 比较），则挪LEFT  往右端的深沟走
time: o(logn) worst o(n)
space: o(1)

 1 public int findMin(int[] nums) {
 2         if (nums ==null || nums.length ==0) return -1 ;
 3         int left = 0, right = nums.length -1 ;
 4         while(left + 1 < right){
 5             int mid = left + (right - left)/2 ;
 6             if (nums[left] == nums[mid]) {
 7                 left++;
 8             }
 9             if (nums[right] == nums[mid]) {
10                 right--;
11             }
12             //the lower part
13             if (nums[mid]<nums[right]) {
14                 right = mid;
15             }
16             //the upper part
17             else if(nums[mid]>nums[right]){
18                 left = mid ;
19             }
20         }
21         //post processing
22         return Math.min(nums[left], nums[right]);
23     }

注意针对 rotated sorted array 去重复的固定套路

参考 33,81,153, 154