【Hot100】LeetCode—153. 寻找旋转排序数组中的最小值

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• 原题链接：153. 寻找旋转排序数组中的最小值

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1- 思路

二分

• 左区间二分找分界点，二分找到旋转后的分界点即可
• 以 nums[mid] 为基准，对比 nums[0] 即可找到区间分界点

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2- 实现

⭐33. 搜索旋转排序数组——题解思路

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

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3- ACM 实现

public class findMin {

    public static int binaryS(int[] nums){
        int left= 0;
        int right = nums.length-1;
        while(left<right){
            int mid = (left+right)/2+1;
            if(nums[mid] >= nums[0]){
                left = mid;
            }else {
                right = mid-1;
            }
        }
        if(right == nums.length-1) return nums[0];
        return nums[right+1];
    }


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String input = sc.nextLine();
        input = input.replace("[","").replace("]","");
        String[] parts = input.split(",");
        int[] nums = new int[parts.length];
        for(int i = 0 ; i < nums.length; i++){
            nums[i] = Integer.parseInt(parts[i]);
        }
        System.out.println("结果是"+binaryS(nums));
    }
}

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