# Third Maximum Number with Java

## The challenge

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:1Explanation:The third maximum is 1.

**Example 2:**

Input:[1, 2]Output:2Explanation:The third maximum does not exist, so the maximum (2) is returned instead.

**Example 3:**

Input:[2, 2, 3, 1]Output:1Explanation:Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum.

## The solution

I have seen many possible resolutions to this problem, but my favourite has to be the following:

```
class Solution {
// input a primitive array of ints and return an int
public int thirdMax(int[] nums) {
// set variables for our 3 max'es
Integer max1 = null;
Integer max2 = null;
Integer max3 = null;
// loop through the input array
for (Integer n : nums) {
// if already set, then skip this loop iteration
if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
// if 1
if (max1 == null || n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
// if 2
} else if (max2 == null || n > max2) {
max3 = max2;
max2 = n;
// if 3
} else if (max3 == null || n > max3) {
max3 = n;
}
}
// return max3, unless null
return max3 == null ? max1 : max3;
}
}
```