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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.
代码如下：
1 public class Solution { 2 public boolean increasingTriplet(int[] nums) { 3 4 if(nums.length<3) 5 return false; 6 7 for(int i=nums.length-1;i>=2;i--) 8 { 9 int j=i-1; 10 for( ;j>=1;j--) 11 { 12 if(nums[i]>nums[j]) 13 { 14 int k=j-1; 15 for(;k>=0;k--) 16 { 17 if(nums[j]>nums[k]) 18 return true; 19 } 20 } 21 22 } 23 24 } 25 return false; 26 } 27 }
334. Increasing Triplet Subsequence
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原文地址：http://www.cnblogs.com/ghuosaao/p/5421451.html
