LeetCode No334. Increasing Triplet Subsequence

1. 问题描述

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.

2. 思路

1. 从头开始遍历，并用min记录最小值，secondmin记录次最小值
2. 当min和secondmin都找到时，只要存在一个新的值大于这两个值，那么就存在递增的三元子串。
3. 特别的，例如[1, 2, 0, 4] 在遍历时，min可能会变成0,但是没关系，min和secondmin的最大值仍为secondmin，依然可以用于判断，且之前存在oldmin(先前的min)和secondmin之间的递增序列。

3. 代码

class Solution {public:    bool increasingTriplet(vector<int>& nums) {        int min = INT_MAX, secondmin = INT_MAX;        for (auto i : nums) {            if (i <= min)                min = i;            else if (i <= secondmin)                secondmin = i;            else                 return true;        }        return false;    }};

4. 参考资料

1. https://leetcode.com/discuss/86593/clean-and-short-with-comments-c