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1. 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.
2. Answer
solution1 :
public class Solution { public boolean increasingTriplet(int[] nums) { for (int i = 0; i < nums.length - 2; i++) { for (int j = i + 1; j < nums.length - 1; j++) { if (nums[j] > nums[i]) { for (int k = j + 1; k < nums.length; k++) { if (nums[k] > nums[j]) return true; } } } } return false; } }
solution2：
public class Solution { public boolean increasingTriplet(int[] nums) { int min = Integer.MAX_VALUE; int secondMin = Integer.MAX_VALUE; for (int i = 0; i < nums.length; i++){ if (nums[i] <= min){ min = nums[i]; } else if (nums[i] <= secondMin){ secondMin = nums[i]; } else { return true; } } return false; } }
solution2 has a better time complexity, it is a better way.
【LeetCode】Increasing Triplet Subsequence（334）
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原文地址：http://www.cnblogs.com/leesf456/p/5354452.html
