Description:
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:
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.
Link:
https://leetcode.com/problems/third-maximum-number/#/description
解题方法：
用三个数分别储存第一、第二、第三大的数，当出现等于这三个数的情况，不进行更新。
Tips:
为了防止数组中有INT_MIN，用long来储存，这样初始化为比INT_MIN还小的数，最后就可以知道第三大的数有没有更新。
Time Complexity：
O(N)
完整代码：
int thirdMax(vector<int>& nums)
{
long max1 = (long)INT_MIN - 1, max2 = (long)INT_MIN - 1, max3 = (long)INT_MIN - 1;
for(int i: nums)
{
if(i == max1 || i == max2 || i == max3)
continue;
if(i > max1)
{
max3 = max2;
max2 = max1;
max1 = i;
}
else
if(i > max2)
{
max3 = max2;
max2 = i;
}
else
if(i > max3)
max3 = i;
}
return max3 == (long)INT_MIN - 1 ? max1 : max3;
}
