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leetcode(581):Shortest Unsorted Continuous Subarray

程序员文章站 2022-03-07 18:28:07
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题目

最短需要排序的子序列

  1. Shortest Unsorted Continuous Subarray
    Easy

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 <=.

思路

关键是找到6,4和10,9
6.4是从右到左最后一对大小组合
10.9是从左到右最后一对小大组合
这件事一开始不是很好理解,大家多想想。
leetcode(581):Shortest Unsorted Continuous Subarray

代码

	class Solution {
	public:
		int findUnsortedSubarray(vector<int>& nums) {
			int begin = 0;
			int end = nums.size() - 1;
			int max = nums[0];
			int min = nums[end];
			for (int i = 0; i < nums.size(); i++) {
				if (nums[i] >= max) {
					max = nums[i];
				}
				else
					begin = i;
			
			}


			for (int i = nums.size() - 1; i>=0; i--) {
				if (nums[i] <= min) {
					min = nums[i];

				}
				else
					end = i;


			}

			if (end < begin)
				return begin- end + 1;
			else
				return 0;
		}
	};