leetcode笔记:Increasing Triplet Subsequence
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一. 题目描述
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.
二. 题目分析
题目大意是,给定一个无序数组,判断其中是否存在一个长度为3的递增子序列。
即是,如果存在下标i, j, k(0 ≤ i < j < k ≤ n-1)
,使得arr[i] < arr[j] < arr[k]
,返回true
,否则返回false
。
要求满足O(n)
的时间复杂度和O(1)
的空间复杂度。
应该注意到,题目要求只要在数组中找到三个递增的元素即可,不要求这三个元素是否连续,因此,只需维护两个整数变量a, b
,用来记录数组中大小递增的前2
个元素,满足条件时,应该有:a < b < nums[i]
。
三. 示例代码
class Solution {public: bool increasingTriplet(vector<int>& nums) { int n = nums.size(); if (n < 3) return false; int a = INT_MAX, b = INT_MAX; for (int i = 0; i < n; ++i) { if (nums[i] <= a) a = nums[i]; else if (nums[i] <= b) b = nums[i]; else return true; } return false; }};
四. 小结
在搜索过程中,需尽量使a
和b
变小,同时需保证a
小于b
。
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