581. Shortest Unsorted Continuous Subarray（python+cpp）

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本文介绍了一种高效算法，用于找出使整个数组有序所需的最短连续子数组长度。通过对比原数组与排序后的数组，使用双指针技术确定未排序部分的边界。

题目：

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 <=.

解释：
先把原数组排序以后，双指针诸葛比较即可。
python代码：

class Solution(object):
    def findUnsortedSubarray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        sorted_nums=sorted(nums)
        if sorted_nums==nums:
            return 0  
        left=0
        n=len(nums)
        right=n-1
        while left<n and nums[left]==sorted_nums[left]:
            left+=1
        while right>=0 and nums[right]==sorted_nums[right]:
            right-=1
        return right-left+1

c++代码：

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