【LeetCode】Increasing Triplet Subsequence（334）

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.