414. Third Maximum Number

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.

解题思路为采用三个变量来记录最大，次大和第三大的数。

注意点是：

1. 采用Integer，这样可以判断是否存在第三大的值；
2. 重复的值需要处理。

public class Solution {
    public int thirdMax(int[] nums) {
        Integer max1 = null;
        Integer max2 = null;  
        Integer max3 = null;
        for(Integer n : nums) {
            if(n.equals(max1) || n.equals(max2) || n.equals(max3)) continue; // skip duplicates
            if(max1 == null || n > max1) {
                max3 = max2;
                max2 = max1;
                max1 = n;
            } else if(max2 == null || n > max2) {
                max3 = max2;
                max2 = n;
            } else if(max3 == null || n > max3) {
                max3 = n;
            }
        }
        return max3 == null ? max1 : max3;
    }
}