Question
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.
Code
public class Solution {
public boolean increasingTriplet(int[] nums) {
// if (nums == null || nums.length < 3) return false;
// int[] dp = new int[nums.length];
// dp[0] = 1;
// for (int i = 1; i < dp.length; i++) {
// dp[i] = 1;
// for (int j = 0; j < i; j++) {
// if (nums[j] < nums[i]) {
// dp[i] = Math.max(dp[i], dp[j] + 1);
// }
// if (dp[i] == 3) return true;
// }
// }
// return false;
int min = Integer.MAX_VALUE, secondMin = Integer.MAX_VALUE;
for (int n: nums) {
if (n <= min) {
min = n;
} else if (n <= secondMin) {
secondMin = n;
} else return true;
}
return false;
}
}
Solution
一開始想用類似最長遞增子序列的dp思路，雖未超時(shí)但是時(shí)間復(fù)雜度不符合要求痢法。
用兩個(gè)int分別記錄最小值和次小值衷咽，如果有值比這兩個(gè)值大载矿，說明存在類似的三元組。
