# 题目描述：

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.

# 代码实现：

``````class Solution {
public:
int thirdMax(vector<int>& nums) {
long first=LONG_MIN,second=LONG_MIN,third=LONG_MIN;
for(int i=0;i<nums.size();i++){
if(nums[i]>first){
third = second;
second = first;
first = nums[i];
}
else if(nums[i]<first&&nums[i]>second){
third = second;
second = nums[i];
}
else if(nums[i]<second&&nums[i]>third){
third = nums[i];
}
}
return (third==LONG_MIN||third==second)?first:third;
}
};
``````