334. Increasing Triplet Subsequence - jiejackyzhang/leetcode-note GitHub Wiki

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.

解题思路为记录两个最小值，min和secondMin。 顺序扫描数组，若num均大于min和secondMin，则返回true。 这里需要注意的是，min和secondMin不一定就是该triplet。 因为更新时，min可能会在secondMin之后，但是此时secondMin前面必然有一个比它小的，因此num>secondMin即满足条件。

public class Solution {
    public boolean increasingTriplet(int[] nums) {
        int min = Integer.MAX_VALUE, secondMin = Integer.MAX_VALUE;
        for(int num : nums) {
            if(num <= min) {
                min = num;
            } else if(num <= secondMin) {
                secondMin = num;
            } else {
                return true;
            }
        }
        return false;
    }
}