LeetCode challenge – Third Maximum Number – Java Solution
This is a solution for the LeetCode challenge – Third Maximum Number written in Java ( Source )
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.
public class ThirdMaximumNumber {
public static void main(String[] args) throws Exception {
int[] a = {1, 2};
System.out.println(new ThirdMaximumNumber().thirdMax(a));
}
public int thirdMax(int[] nums) {
long[] max = {Long.MIN_VALUE, Long.MIN_VALUE, Long.MIN_VALUE};
int count = 0;
for (int num : nums) {
for (int j = 0; j < 3; j++) {
if (max[j] > num) continue;
else if (max[j] == num) break;
int k = j;
long temp1, temp2;
temp1 = num;
count++;
while (k < 3) {
temp2 = max[k];
max[k] = temp1;
temp1 = temp2;
k++;
}
break;
}
}
System.out.println(Integer.MIN_VALUE);
return (count >= 3) ? (int) max[2] : (int) max[0];
}
}
