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.

The array may contain duplicates.

一刷

public class Solution {
    public int findMin(int[] nums) {
        int lo = 0, hi = nums.length-1;
        while(lo<hi){
            if(nums[lo]<nums[hi]) return nums[lo];
            int mid = lo+(hi-lo)/2;
            if(nums[mid]>nums[hi]){//left half is sorted
                lo = mid+1;
            }
            else if(nums[mid]<nums[hi]){//right half is sorted
                hi = mid;
            }
            else hi--;
        }
        return nums[lo];
    }
}

二刷
由于是Rotated Sorted Array, 并且最低点在lo和hi之间，那么lo和hi之间只可能是

所以当noms[lo]<nums[hi]的时候，说明lo就是最低点。
当nums[mid]==nums[lo]的时候，并没有其它信息，只能将lo++， 因为lo和mid相等，可以缩短一个查找范围（mid仍在范围中）

public class Solution {
    public int findMin(int[] nums) {
        int lo = 0, hi = nums.length-1;
        while(lo<hi){
            int mid = lo + (hi-lo)/2;
            if(nums[lo]<nums[hi]) return nums[lo];
            if(nums[mid]>nums[lo]){//on the right side
                lo = mid+1;
            }
            else if(nums[mid]<nums[lo]){
                hi = mid;//on the left side;
            }
            else lo++;
        }
        return nums[lo];
    }
}