LeetCode-334. Increasing Triplet Subsequence
Description:
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
.
题目大意是判断所给的数组中是否存在长度为3的递增序列。
最常规的办法应该是两层循环,逐个判断。时间复杂度较高。还有一种比较巧妙的办法,维护两个距离最近的递增序列,如果能找到第三个则返回true。设置一个small,一个big,small<big。如果能找到一个不小于small和big的就返回true。时间复杂度为O(n)
实现代码:
public class Solution { public boolean increasingTriplet(int[] nums) { if(nums == null || nums.length < 3) return false; int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE; for(int value : nums) { if(value <= small) small = value; else if(value <= big) big = value; else return true; } return false; } }
作者:Pickle
声明:对于转载分享我是没有意见的,出于对博客园社区和作者的尊重一定要保留原文地址哈。
致读者:坚持写博客不容易,写高质量博客更难,我也在不断的学习和进步,希望和所有同路人一道用技术来改变生活。觉得有点用就点个赞哈。
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