414. 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.

题目含义：找出数组中第三大的数字，相同的数字看做是一个。如果不存在第三大的，就返回最大的数字
方法一:

    public int thirdMax(int[] nums) {
        Integer max1 = null;
        Integer max2 = null;
        Integer max3 = null;
        for (Integer n:nums)
        {
            if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
            if (max1 == null || n > max1)
            {
                max3 = max2;
                max2 = max1;
                max1 = n;
            }else if (max2 == null || n > max2)
            {
                max3 = max2;  
                max2 = n;
            }else if (max3 == null || n > max3)
            {
                max3 = n;
            }
        }
        return max3 == null?max1:max3;        
    }

 

方法二：

    public int thirdMax(int[] nums) {
        if (nums.length==0)  return 0;
        Set<Integer> temp =new TreeSet<Integer>(new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o1.compareTo(o2);
            }
        });
        for (Integer number:nums) temp.add(number);
        List<Integer> values =  new ArrayList<>(temp);
        return values.size()<3?values.get(values.size()-1):values.get(values.size()-3);
    }