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:
用一个 HashSet 来判断元素是否曾经出现过，用 PriorityQueue来保存当前最大的3个元素（大小设为4防止 capacity 扩充引发额外开销）
public class Solution
{
public int thirdMax(int[] nums)
{
PriorityQueue<Integer> pq = new PriorityQueue<>(4);
Set<Integer> set = new HashSet<>();
for(int i = 0; i < nums.length; i++)
{
if(!set.contains(nums[i]))
{
set.add(nums[i]);
pq.offer(nums[i]);
if (pq.size() > 3)
pq.poll();
}
}
if(pq.size() > 2)
{
return pq.poll();
}
else
{
int result = 0;
while(!pq.isEmpty())
{
result = pq.poll();
}
return result;
}
}
}
