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).

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. 注意，重复的数字不算。 这题本没啥难度，一个小tricky 就是用Integer 来初始化 max1, max2, ,max3, 因为Integer 可以被初始化成null 而不是 Integer.MIN_VALUE, 特别在比大小时候null 比 Integer.MIN_VALUE 有价值。

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