leetcode 581. Shortest Unsorted Continuous Subarray 最短乱序子数组

ゝ一纸荒年。 2022-06-03 04:40 196阅读 0赞

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 <=.

本题题意很简单，我的做法更简单，直接排序然后比较即可

应该还有更好地方法

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    
    using namespace std;
    
    
    class Solution 
    {
    public:
        int findUnsortedSubarray(vector<int>& nums) 
        {
            int n = nums.size(), i = 0, j = n - 1;
            vector<int> t = nums;
            sort(t.begin(), t.end());
    
            while (i < n && nums[i] == t[i])
                ++i;
            while (j > i && nums[j] == t[j])
                --j;
            return j - i + 1;
        }
    };