[WXM] LeetCode 334. Increasing Triplet Subsequence C++

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

Example 1:

Input: [1,2,3,4,5]
Output: true

Example 2:

Input: [5,4,3,2,1]
Output: false

Approach

1. 题目大意找到长度至少为3的递增子序列，然后题目要求时间复杂度控制在O(n)的级别，空间复杂度控制在O(1)的级别。这道题与求递增子序列的长度的题目很相似，我们可以参照求递增子序列dp的思路，我们用一个first一直记录最小值，然后再用second记录比first大的最小值，这样我们就可以直接找是否有比second大的值就可以了，就可以凑成最短的递增子序列。

Code

class Solution {
public:
	bool increasingTriplet(vector& nums) {
		if (nums.size() < 3)return false;
		int first = INT_MAX, second = INT_MAX;
		for (int i = 2; i < nums.size(); i++) {
			first = min(first, nums[i - 2]);
			if (nums[i - 1] > first) {
				second = min(second, nums[i - 1]);
			}
			if (nums[i] > second)return true;
		}
		return false;
	}
};