LeetCode Third Maximum Number

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.

题解：这道题是要在给定的数组中找出第3大的数，而且这个数组存在重复数据，并且时间复杂度要在o(n)之内。那么就考虑用空间换时间，也就是用一个set，而且还是treeset来存储数组中的数，因为set可以自动帮我们去重；另一方面我采用treeset，是可以帮我自动排序，按照升序排序，所以也非常方便。那么分两种情况考虑，如果是set中的值小于3，那么就认为是不存在第3大的数，考虑需要输出的数组中最大的数；反之，则输出第三大的数。

public static int thirdMax(int[] nums)
{
int length = nums.length;
Set<Integer> set = new TreeSet<>();
for(int i = 0; i < length; i++)
set.add(nums[i]);
System.out.println(set.size());
int target = 0;
if(set.size() < 3)
{
Iterator<Integer> iter = set.iterator();
while(iter.hasNext())
{
target = iter.next();
}
}
else
{
int size = set.size();
int current = 0;
Iterator<Integer> it = set.iterator();
while(it.hasNext())
{
target = it.next();
current += 1;
if(current >= set.size() - 2 )
break;
}
}
return target;
}