【LeetCode】581. Shortest Unsorted Continuous Subarray

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581. Shortest Unsorted Continuous Subarray

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.
Difficulty：Easy，个人感觉难度 Medium
Example:

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 <=.

方法1：暴力求解，超时

• Time complexity : $O\left ( n^{3} \right )$
• Space complexity : $O\left ( 1 \right )$
  思路：
  1.考虑每个子数组nums[i:j]，求出子数组的max和min
  2.假设nums[0:i-1]和nums[j:n-1]是有序的，nums[i:j]是乱序的，需要满足nums[0:i-1]的元素都小于min以及nums[j:n-1]的元素都大于max，此时记录j-i
  3.取最小的j-i

class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
		int n = nums.size();
		int res = n;
		for (int i = 0; i < n; i++) {
			for (int j = i; j <= n; j++) {
				int mins = INT_MAX, maxs = INT_MIN, prev = INT_MIN;
				for (int k = i; k < j; k++) {
					mins = min(mins, nums[k]);
					maxs = max(maxs, nums[k]);
				}
				if ((i > 0 && nums[i - 1] > mins) || (j < n && nums[j] < maxs)) 
					continue;
				int k = 0;
				while (k < i && prev <= nums[k]) {
					prev = nums[k];
					k++;
				}
				if (k != i)
					continue;
				k = j;
				while (k < n && prev <= nums[k]) {
					prev = nums[k];
					k++;
				}
				if (k == n) 
					res = min(res, j - i);
			}
		}
		return res;
    }
};

方法2：两两比较，差解

• Time complexity : $O\left ( n^{2} \right )$
• Space complexity : $O\left ( 1 \right )$
  思路：
  1.每两个元素进行比较，例如nums[i]和nums[j]，i<j<n，如果nums[j]>nums[i]说明这两个元素乱序，记录下左边界和右边界
  2.更新边界时，取最左和最右

class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
		int n = nums.size();
		int left = n, right = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (nums[i] > nums[j]) {
					left = min(i, left);
					right = max(j, right);
				}
			}
		}
		return right - left < 0 ? 0 : right - left + 1;
    }
};