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.
int thirdMax(vector<int>& nums) { // time: O(n); space: O(1)
long mx1 = numeric_limits<long>::min(), mx2 = numeric_limits<long>::min(), mx3 = numeric_limits<long>::min();
for (int num : nums) {
if (mx1 == num || mx2 == num || mx3 == num) continue;
if (num > mx1) {
mx3 = mx2;
mx2 = mx1;
mx1 = num;
} else if (num > mx2) {
mx3 = mx2;
mx2 = num;
} else if (num > mx3) {
mx3 = num;
}
}
return mx3 != numeric_limits<long>::min() ? mx3 : mx1;
}