Sword refers to Offer—— 11. Rotate the smallest number of the array

The topic is as follows

Moving several elements from the beginning of an array to the end of the array is called the rotation of the array. Input a rotation of an ascendingly sorted array, and output the smallest element of the rotated array. For example, the array [3,4,5,1,2] is a rotation of [1,2,3,4,5], and the minimum value of the array is 1.

Example 1:
Input: [3,4,5,1,2]
Output: 1

Example 2:
Input: [2,2,2,0,1]
Output: 0

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Own algorithm and code

(The code can be run and submitted correctly) The
specific algorithm is as follows:

• Initialize the minimum value min=-1;
• Through the array, if satisfied numbers[i] > numbers[i+1], will be numbers[i+1]assigned to min, and out of the loop;
• Determining minwhether or not equal to the initial value -1, i.e., whether the array satisfying numbers[i] > numbers[i+1]conditions;
• If min==-1, there may be two situations: the minimum is the first or the last, so compare the two numbers;
  • If numbers[0] < numbers[numbers.length-1], then min = numbers[0];
  • If numbers[0] > numbers[numbers.length-1], then min = numbers[numbers.length-1];

public class Solution {
    
    

    public static int minArray(int[] numbers) {
    
    
        int min = -1;
        for (int i = 0; i < numbers.length-1; i++){
    
    
            if (numbers[i] > numbers[i+1]){
    
    
                min = numbers[i+1];
                break;
            }
        }
        if (min == -1){
    
    
            if (numbers[0] < numbers[numbers.length-1]){
    
    
                min = numbers[0];
            }else {
    
    
                min = numbers[numbers.length-1];
            }
        }
        return min;
    }

    public static void main(String[] args) {
    
    
        int[] numbers = {
    
    2,3,4,5,1};
        int result = minArray(numbers);
        System.out.println(result);
    }
}

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Reference algorithm and code (binary search method)

class Solution {
    
    
    public int minArray(int[] numbers) {
    
    
        int low = 0;
        int high = numbers.length - 1;
        while (low < high) {
    
    
            int middle = low + (high - low) / 2;
            if (numbers[middle] < numbers[high]) {
    
    
                high = middle ;
            } else if (numbers[middle] > numbers[high]) {
    
    
                low = middle + 1;
            } else {
    
    
                high -= 1;
            }
        }
        return numbers[low];
    }
}