581. Shortest Unsorted Continuous Subarray

原创于 2018-12-19 21:58:48 发布 · 250 阅读

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本文介绍了一种算法，用于在给定整数数组中找到一个连续的子数组，仅需对该子数组进行升序排序即可使整个数组有序。通过一次排序操作和双指针策略，该算法能高效地确定最短的需要排序的子数组长度。

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：

1，如果数组长度<=1，则直接返回0

2，排序

设置两个下标，一个从头开始加，一个从尾开始减，与排序好的数组比较

当两个下标遇到不相等时，则停止更新下标

#pragma once
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

class Solution {
public:
	int findUnsortedSubarray(vector<int>& nums) {
		vector<int> order = nums;
		sort(order.begin(), order.end());
		if (nums.size() == 1)
		{
			return 0;
		}
		int i = 0, j = nums.size() - 1;
		
		for (int k = 0; i < nums.size()&&j>=i&&k<nums.size();k++)
		{
			//从头开始
			if (i == 0 && nums[i] == order[0])//第一个数从小到大
			{
				i++;
			}
			else if(nums[i] == order[i])
			{
				i++;
			}
			
			//从尾开始
			if (nums[j] == order[j]&& j == nums.size() - 1)//最后一个数，从小到大
			{
				j--;
			}
			else if (nums[j] == order[j])
			{
				j--;
			}
			

			//相等则为已经排好序的
			if (i == j)
			{
				return 0;
			}
		
		}
		return j - i + 1;

	}
};