【leetcode】153. Find the lowest value in the rotation sort array

subject ：
153. Look for the minimum value in the rotation sort array
We know a length of n Array of , In ascending order in advance , Through 1 To n Time rotate after , Get the input array . for example , Original array nums = [0,1,2,4,5,6,7] After the change may get ：
If you rotate 4 Time , You can get [4,5,6,7,0,1,2]
If you rotate 7 Time , You can get [0,1,2,4,5,6,7]
Be careful , Array [a[0], a[1], a[2], …, a[n-1]] Rotate once The result is an array [a[n-1], a[0], a[1], a[2], …, a[n-2]] .

Give you an element value Different from each other Array of nums , It turns out to be an ascending array , According to the above situation, we have made many rotations . Please find and return the The smallest element .

Example 1：

Input ：nums = [3,4,5,1,2]
Output ：1
explain ： The original array is [1,2,3,4,5] , rotate 3 Get the input array for the first time .
Example 2：

Input ：nums = [4,5,6,7,0,1,2]
Output ：0
explain ： The original array is [0,1,2,4,5,6,7] , rotate 4 Get the input array for the first time .
Example 3：

Input ：nums = [11,13,15,17]
Output ：11
explain ： The original array is [11,13,15,17] , rotate 4 Get the input array for the first time .

Dichotomy ：
Because the array does not contain duplicate elements , And as long as the current interval length is not 1, mid It won't be with high coincidence ; If the current interval length is 1, This shows that we can end the two-point search .

When the binary search ends （l=r), We get the position of the minimum .

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