581. Shortest Unsorted Continuous Subarray (LeetCode)-优快云博客

本文介绍了一种算法，用于找出一个整数数组中需要排序的最短连续子数组长度，以使整个数组按升序排列。通过一个示例解释了如何确定需要排序的子数组，并提供了一个Java实现解决方案。

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Description:

Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.

You need to find the shortest such subarray and output its length.

Example 1:

Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

Note:

Then length of the input array is in range [1, 10,000].
The input array may contain duplicates, so ascending order here means <=.

Accepted

80,993

Submissions

265,634

Solution:

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        
        int [] tmp = new int[nums.length];
        
        for(int i = 0; i < nums.length; i++){
            
            tmp[i] = nums[i];
        }
        
        int start = 0;
        int end = 0;
        Arrays.sort(nums);

          for(int i = 0; i<nums.length; i++){
            
            if(nums[i]!=tmp[i]){
                start = i;
                break;
            }
        }
         
            for(int k = nums.length-1; k>=0; k--){
            
            if(nums[k]!=tmp[k]){
                end = k;
                break;
            }
        }
        // System.out.println("Start "+start);
         // System.out.println("end "+end);
        if(start ==end && start ==0){
            return 0;
        }
        return end-start+1;
    }
}