LeetCode – 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.
这道题的test case写的很巧妙

public class Solution {
    public int thirdMax(int[] nums) {
        Integer f = null;
        Integer s = null;
        Integer result = null;
        for(int i=0;i<nums.length;i++){
            System.out.println(nums[i]);
            if((f != null && nums[i] ==f) || (s != null && nums[i] ==s) || (result != null && nums[i] ==result)){
                continue;
            }
            if(f == null || nums[i]>f){
                result = s;
                s = f;
                f = nums[i];
            }
            else if(s == null || nums[i]>s){
                result = s;
                s = nums[i];
            }
            else if(result == null || nums[i]>result){
                result = nums[i];
            }
        }
        if(nums.length <3 || result == null){
            return f;
        }
        return result;
    }
}

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