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

本文链接：https://blog.youkuaiyun.com/weixin_44610959/article/details/108888060

此篇博客介绍如何通过一次遍历找到只需对一连续子数组进行升序排列即可使整个数组有序的最短长度。通过实例讲解Solution类中的findUnsortedSubarray方法，适用于长度在1到10,000的整数数组，且可能包含重复元素。

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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 <=.

难度：简单

class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        int len =nums.size();
        int num_min=INT_MAX;
        int num_max=INT_MIN;
        //一次扫描，同时求出无序子数组中最小元素和最大元素
        for(int i=0;i<len-1;i++){
            if(nums[i+1]<nums[i]){
                num_min=min(nums[i+1],num_min);
                num_max=max(nums[i],num_max);
            }
        }
        //确定最小元素的正确位置
        int l=0;
        for(int i=0;i<len-1;i++){
            if(nums[i]>num_min) {
                l=i;
                break;
            }
        }
        //确定最大元素的正确位置
        int r=0;
        for(int i=len-1;i>-1;i--){
            if(nums[i]<num_max){
                 r=i;
                 break;
            }
        }
        //返回结果这里需要特殊处理一下数组已是升序如[1,2,3,4]，l=0，r=0的特殊情况
        if(!r&&!l) return 0;
        else return r-l+1;

    }
};

ascending	递增