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.

本题注意点是nums数组不是空集，本题考查Integer的特点，值得注意的是其equals的方法：比较此对象与指定对象。当且仅当参数不为 null，并且是一个与该对象包含相同 int 值的 Integer 对象时，结果为 true。
代码如下：

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;
            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;
    }
}