# LeetCode challenge – Third Maximum Number – Java Solution

This is a solution for the LeetCode challenge – Third Maximum Number written in Java ( Source )

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).*Example 1: Input: [3, 2, 1]Output: 1*

Explanation: The third maximum is 1. Example 2:

*Input: [1, 2]Output: 2*

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

*Input: [2, 2, 3, 1]Output: 1*

Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum.

public class ThirdMaximumNumber { public static void main(String[] args) throws Exception { int[] a = {1, 2}; System.out.println(new ThirdMaximumNumber().thirdMax(a)); } public int thirdMax(int[] nums) { long[] max = {Long.MIN_VALUE, Long.MIN_VALUE, Long.MIN_VALUE}; int count = 0; for (int num : nums) { for (int j = 0; j < 3; j++) { if (max[j] > num) continue; else if (max[j] == num) break; int k = j; long temp1, temp2; temp1 = num; count++; while (k < 3) { temp2 = max[k]; max[k] = temp1; temp1 = temp2; k++; } break; } } System.out.println(Integer.MIN_VALUE); return (count >= 3) ? (int) max[2] : (int) max[0]; } }