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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];
  }
}

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