LeetCode 334. Increasing Triplet Subsequence

Posted 2022-09-25 dylan-java-nyc

原题链接在这里：https://leetcode.com/problems/increasing-triplet-subsequence/

题目：

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.

Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

Input: [1,2,3,4,5]
Output: true

Example 2:

Input: [5,4,3,2,1]
Output: false

题解：

Could use dp to track up to index i, the length of longest subarray. 

For all j from 0 to i, if nums[j]<nums[i], then dp[i] = Math.max(dp[i], dp[j]+1). If dp[i] >= 3, then return true.

Time Complexity: O(n^2). n = nums.length.

Space: O(n).

AC Java: 

 1 class Solution 
 2     public boolean increasingTriplet(int[] nums) 
 3         if(nums == null || nums.length < 3)
 4             return false;
 5         
 6         
 7         int len = nums.length;
 8         int [] dp = new int[len];
 9         for(int i = 0; i<len; i++)
10             dp[i] = 1;
11             for(int j = 0; j<i; j++)
12                 if(nums[j]<nums[i])
13                     dp[i] = Math.max(dp[i], dp[j]+1);
14                     if(dp[i] > 2)
15                         return true;
16                     
17                 
18             
19         
20         
21         return false;
22     
23 

Could use first and second variables to maintain the first smallest and second smallest number.

For each number in nums array, if num <= first, update first. Else if num <= second, update second. Else, this means that there are 2 increasing numbers smaller than current number before. Thus return true.

Time Complexity: O(n).

Space: O(1).

AC Java:

 1 class Solution 
 2     public boolean increasingTriplet(int[] nums) 
 3         int first = Integer.MAX_VALUE;
 4         int second = Integer.MAX_VALUE;
 5         for(int num : nums)
 6             if(num <= first)
 7                 first = num;
 8             else if(num <= second)
 9                 second = num;
10             else
11                 return true;
12             
13         
14         
15         return false;
16     
17 

类似Longest Increasing Subsequence.