【LeetCode】Increasing Triplet Subsequence(334)

时间:2023-01-28 04:47:57

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 < kn-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.