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.
Solution:
public class Solution {
public int thirdMax(int[] nums) {
Integer first=null, second=null, third=null;
for(Integer n : nums){
if(n.equals(first) || n.equals(second) || n.equals(third)) continue;
if(first==null || n>first){
third = second;
second = first;
first = n;
} else if(second==null || n>second){
third = second;
second = n;
} else if(third==null || n>third){
third = n;
}
}
return third==null ? first : third;
}
}