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.

My Solution:

public class Solution {
    public int thirdMax(int[] nums) {
        long fstMax = Long.MIN_VALUE;
        long sndMax = Long.MIN_VALUE;
        long thirdMax = Long.MIN_VALUE;
        for (long each : nums) {
            if (each > fstMax) {
                long temp = fstMax;
                fstMax = each;
                each = temp;
            }
            if (each<fstMax && each>sndMax) {
                long temp = sndMax;
                sndMax = each;
                each = temp;
            }
            if (each<fstMax && each<sndMax && each>thirdMax) {
                thirdMax = each;
            }
        }
        return thirdMax==Long.MIN_VALUE? (int)fstMax : (int)thirdMax;
    }
}

Highest votes in discussion

public int thirdMax(int[] nums) {
        int max, mid, small, count;
        max = mid = small = Integer.MIN_VALUE;
        count = 0;  //Count how many top elements have been found.

        for( int x: nums) {
            //Skip loop if max or mid elements are duplicate. The purpose is for avoiding right shift.
            if( x == max || x == mid ) {
                continue;
            }

            if (x > max) {
                //right shift
                small = mid;
                mid = max;

                max = x;
                count++;
            } else if( x > mid) {
                //right shift
                small = mid;

                mid = x;
                count++;
            } else if ( x >= small) { //if small duplicated, that's find, there's no shift and need to increase count.
                small = x;
                count++;
            }
        }

        //"count" is used for checking whether found top 3 maximum elements.
        if( count >= 3) { 
            return small;
        } else {
            return max;
        }
    }