334. Increasing Triplet Subsequence
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?
class Solution {
public boolean increasingTriplet(int[] nums) {
if(nums == null || nums.length < 3) {
return false;
}
Integer num1 = nums[0], num2 = null;
for(int i = 1; i < nums.length; i++){
if(nums[i] <= num1){
num1 = nums[i];
}else{
if(num2 != null && nums[i] > num2){
return true;
}
num2 = nums[i];
}
}
return false;
}
}