334. Increasing Triplet Subsequence

Medium

Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.

Example 1:

Input: nums = [1,2,3,4,5]
Output:
 true
Explanation:
 Any triplet where i < j < k is valid.

Example 2:

Input: nums = [5,4,3,2,1]
Output:
 false
Explanation:
 No triplet exists.

Example 3:

Input: nums = [2,1,5,0,4,6]
Output:
 true
Explanation:
 The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.

Constraints:

• 1 <= nums.length <= 5 * 105

• -231 <= nums[i] <= 231 - 1

Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

解題

設立兩變數small和medium來記錄最小與中間的兩個數字，而一但找到這兩個數，又找到大於這兩數的第三數，回傳true。

func increasingTriplet(nums []int) bool {
	small, medium := math.MaxInt32, math.MaxInt32

	for _, val := range nums {
		if val <= small {
			small = val
		} else if val <= medium {
			medium = val
		} else {
			return true
		}
	}

	return false
}