Increasing Triplet Subsequence

最新推荐文章于 2022-07-22 03:27:20 发布

zhanghuanzj

最新推荐文章于 2022-07-22 03:27:20 发布

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分类专栏： leetcode 文章标签： c++ leetcode Increasing Triplet S

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本文介绍了一个算法，用于判断一个未排序数组中是否存在长度为3的递增子序列。该算法的时间复杂度为O(n)，空间复杂度为O(1)。通过维护两个变量来记录可能的子序列的前两个元素，并遍历数组进行判断。

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334.Increasing Triplet Subsequence

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

class Solution {
public:
	bool increasingTriplet(vector<int>& nums) {
		int first = INT_MAX;
		int second = INT_MAX;                                              //通过两个变量记录序列的前两个值
		for (size_t i=0;i+1<nums.size();++i)
		{
			if (nums[i]<nums[i+1]&&first>nums[i]&&second>nums[i+1])    //当first与second均比当前值和随后值大时更新
			{
				first = nums[i];
				second = nums[i+1];
			}
			if (nums[i]>second||nums[i+1]>second)                      //若当前值或随后值比second大则形成递增序列
			{
				return true;
			}
			if (nums[i]>first&&nums[i]<second)                         //若当前值处于first和second中间，更新second
			{
				second = nums[i];
			}
		}
		return false;
	}
};