LeetCode #581 - Shortest Unsorted Continuous Subarray

最新推荐文章于 2020-12-09 16:28:59 发布

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本文介绍了一种算法，用于找到整数数组中最短的连续无序子数组，并给出其长度。通过对比原数组与排序后的数组，找出两端相同的元素，从而确定最长无序子数组的位置。

题目描述：

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 <=.

求数组中无序的子数组的最大长度。假设数组中存在最长无序的子数组，那么在它之前的元素都是有序的，并且都小于子数组的最小值，同理在它之后的元素也都是有序的，并且都大于子数组的最大值。所以直接将数组排序然后与原数组比较，除去两头相同的部分就是最长无序子数组。

class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        vector<int> temp=nums;
        sort(temp.begin(),temp.end());
        int begin=0;
        while(begin<nums.size())
        {
            if(nums[begin]==temp[begin]) begin++;
            else break;
        }
        int end=nums.size()-1;
        while(end>=begin)
        {
            if(nums[end]==temp[end]) end--;
            else break;
        }
        return end-begin+1;
    }
};