LeetCode-Increasing Triplet Subsequence

标签：

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.

Analysis:

it is a simplified version of LIS.

Solution:

 1 public class Solution {
 2     public boolean increasingTriplet(int[] nums) {
 3         if (nums.length < 3) return false;
 4         
 5         int p1=0;
 6         int p2 = -1;
 7         for (int i=1;i<nums.length;i++){
 8             if (nums[i] <= nums[p1]){
 9                 p1 = i;
10             } else if (p2==-1 || nums[i] <= nums[p2]){
11                 p2 = i;
12             } else return true;
13         }
14         return false;
15         
16     }
17 }

LeetCode-Increasing Triplet Subsequence

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原文地址：http://www.cnblogs.com/lishiblog/p/5724723.html