leetcode:Increasing Triplet Subsequence

martin_liang

于 2016-03-01 00:46:36 发布

阅读量379

点赞数

CC 4.0 BY-SA版权

分类专栏： C++/C 算法 leetcode

本文链接：https://blog.youkuaiyun.com/martin_liang/article/details/50769472

算法同时被 3 个专栏收录

537 篇文章

订阅专栏

C++/C

504 篇文章

订阅专栏

leetcode

230 篇文章

订阅专栏

本博客探讨如何在给定的未排序数组中，判断是否存在长度为3的递增子序列。通过O(n)的时间复杂度和O(1)的空间复杂度实现算法，并提供了解题思路和代码实现。

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.

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

请参考： https://leetcode.com/discuss/87204/accepted-java-solution-this-question-only-lines-clear-concise

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        
        if (nums.size() < 3)
            return false;
        
        int c1 = INT_MAX;
        int c2 = INT_MAX;
        
        for (int i=0; i<nums.size(); i++)
        {
            if (nums[i] <= c1)
                c1 = nums[i];
            else if (nums[i] <= c2)
                c2 = nums[i];
            else 
                return true;
        }
        return false;
        
    }
};