Increasing Triplet Subsequence

原创于 2017-10-06 14:05:31 发布 · 199 阅读

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本文介绍了一种高效算法，用于检测一个未排序数组中是否存在长度为3的递增子序列。该算法的时间复杂度为O(n)，空间复杂度为O(1)。通过实例展示了算法的应用，如在数组[1,2,3,4,5]中返回true，在数组[5,4,3,2,1]中返回false。

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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.

class Solution {
    public boolean increasingTriplet(int[] nums) {
        if(nums.length<3) return false;
        int first=Integer.MAX_VALUE;
        int second=Integer.MAX_VALUE;
        int third=Integer.MAX_VALUE;
        for(int i=0;i<nums.length;i++){
            if(nums[i]>=third) continue;
            else if(nums[i]>second&&nums[i]<third){
                third=nums[i];
            }else if(nums[i]>first&&nums[i]<=second){
                second=nums[i];
            }else{
                first=nums[i];
            }
        }
        return third<Integer.MAX_VALUE;
    }
}