LC.153.Find Minimum in Rotated Sorted Array

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
如果在左端，则挪LEFT  往右端的深沟走

time: o(logn) space: o(1)

public int findMin(int[] nums) {
        if (nums ==null || nums.length ==0) return -1 ;
        int left = 0, right = nums.length -1 ;
        while(left + 1 < right){
            int mid = left +(right-left)/2 ;
            //lower part
            if (nums[mid]<nums[right]){
                right = mid ;
            }
            //upper part: 这里没有 duplicate, 所以写 nums[mid] > nums[left] 一样通过
            else if(nums[mid]>nums[right]){
                left = mid ;
            }
        }
        //post processing
        return Math.min(nums[left], nums[right]) ;
    }

这个笨方法也能通过

public int findMin(int[] nums) {
        if (nums ==null || nums.length ==0) return -1 ;
        int left = 0, right = nums.length -1 ;
        while(left + 1 < right){
            int mid = left +(right-left)/2 ;
            //this is the case for test case [1,2,3]: not necessary rotated
            if(nums[left] <= nums[mid] && nums[mid]<=nums[right]){
                right = mid ;
            }
            else if (nums[mid] > nums[left] && nums[mid]>=nums[right]){
                left = mid ;
            } else if(nums[mid]<=nums[left]){
                right = mid ;
            }
        }
        //post processing
        return Math.min(nums[left], nums[right]) ;
    }