标签：value time note OLE exit sort tle post input
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.
LeetCode 334. Increasing Triplet Subsequence
标签：value time note OLE exit sort tle post input
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11450797.html
