334. Increasing Triplet Subsequence

最新推荐文章于 2025-01-13 15:16:10 发布

转载最新推荐文章于 2025-01-13 15:16:10 发布 · 130 阅读

本文介绍了一个高效的算法，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)，并提供了具体的实现代码。

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) {
        
        if(nums.size()<=2) return false;
        int c1=INT_MAX;
        int c2=INT_MAX;
        for(auto num:nums)
        {
            if(num<=c1)
                c1 = num;
            else if(num<=c2)
                c2 = num;
            else
                return true;
        }
        return false;
    }
};