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?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
}