leetcode-414-easy

Third Maximum Number

Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.

Example 1:

Input: nums = [3,2,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.
Example 2:

Input: nums = [1,2]
Output: 2
Explanation:
The first distinct maximum is 2.
The second distinct maximum is 1.
The third distinct maximum does not exist, so the maximum (2) is returned instead.
Example 3:

Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2's are counted together since they have the same value).
The third distinct maximum is 1.
Constraints:

1 <= nums.length <= 104
-231 <= nums[i] <= 231 - 1
Follow up: Can you find an O(n) solution?

思路一：用有序 map 记录所有值，在去重的同时做好了排序

    public int thirdMax(int[] nums) {

        Map<Integer, Integer> map = new TreeMap<>();
        Arrays.sort(nums);

        for (int num : nums) {
            map.compute(num, (k, v) -> v == null ? 1 : v + 1);
        }

        Set<Integer> set = map.keySet();
        if (set.size() < 3) {
            return set.toArray(new Integer[]{})[set.size() - 1];
        } else {
            return set.toArray(new Integer[]{})[set.size() - 3];
        }
    }