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
解題方法：
用三個(gè)數(shù)分別儲(chǔ)存第一、第二、第三大的數(shù)型宝，當(dāng)出現(xiàn)等于這三個(gè)數(shù)的情況，不進(jìn)行更新甘有。
Tips:
為了防止數(shù)組中有INT_MIN咧织，用long來(lái)儲(chǔ)存舔痕，這樣初始化為比INT_MIN還小的數(shù)，最后就可以知道第三大的數(shù)有沒(méi)有更新熟空。
Time Complexity：
O(N)
完整代碼：
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;
}
