334. Increasing Triplet Subsequence

最新推荐文章于 2022-07-22 03:27:20 发布

原创最新推荐文章于 2022-07-22 03:27:20 发布 · 620 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

164 篇文章

订阅专栏

本文介绍了一种高效算法，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)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

遍历更新最小的数和次小的数，如果某个数大于这两个数，则有三个数递增，返回true.

class Solution {
public:
	bool increasingTriplet(vector<int>& nums) {
		if (nums.size() < 3) return false;
		int firstMin = INT_MAX;
		int secondMin = INT_MAX;
		for (int i = 0; i < nums.size(); i++){
			if (nums[i] <= firstMin){//<=
				firstMin = nums[i];
			}
			else if (nums[i] <= secondMin){
				secondMin = nums[i];
			}
			else{
				return true;
			}
		}
		return false;
	}
};