414. Third Maximum Number

Problem:

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.

Analysis:

这道题真垃圾，毫无技术含量。有人用一次循环，三个变量来做，头都看晕了，easy的题何必呢。Test case居然有Integer.MIN_VALUE，非得用long转换一下。

Solution:

class Solution {
    public int thirdMax(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        long max = Long.MIN_VALUE, secMax = Long.MIN_VALUE, thirdMax = Long.MIN_VALUE;
        
        for (int i: nums) {
            max = Math.max(i, max);
        }
        for (int i : nums) {
            if (i != max) {
                secMax = Math.max(i, secMax);
            }
        }
        
        for (int i: nums) {
            if (i != max && i != secMax) {
                thirdMax = Math.max(i, thirdMax);
            }
        }
        return thirdMax == Long.MIN_VALUE ? (int)max : (int)thirdMax;
    }
}