Java旋转数组中最小数字具体实现(图文详解版)

public class Solution {
    public int minNumberInRotateArray(int[] nums) {
        if(nums.length == 0){
            return -1;
        }
        int min = nums[0];
        for (int i = 1; i < nums.length; i++) {
            if (min > nums[i]) {
                min = nums[i];
            }
        }
        return min;
    }
}

import java.util.Arrays;
public class Solution {
    public int minNumberInRotateArray (int[] nums) {
        if(nums.length == 0){
            return -1;
        }
        Arrays.sort(nums);
        return nums[0];
    }
}

public class Solution {
    public int minNumberInRotateArray(int[] nums) {
        if(nums.length == 0){
            return -1;
        }
        int left = 0;
        int right = nums.length - 1;
        while(left < right){
            //找到数组的中点
            int mid = (left + right) / 2;
            if(nums[mid] > nums[right]){
                //旋转点在[mid+1, j]中，跳过mid+1左边元素
                left = mid + 1;
            } else if(nums[mid] < nums[right]){
                //旋转点在[left, mid]中，跳过mid右边元素
                right = mid;
            }else{
                //缩小right继续查找
                right--;
            }
        }
        //返回旋转点
        return nums[left];
    }
}