LeetCode_414. Third Maximum Number

414. Third Maximum Number

Easy

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.

package leetcode.easy;

public class ThirdMaximumNumber {
	@org.junit.Test
	public void test() {
		int[] nums1 = { 3, 2, 1 };
		int[] nums2 = { 1, 2 };
		int[] nums3 = { 2, 2, 3, 1 };
		System.out.println(thirdMax(nums1));
		System.out.println(thirdMax(nums2));
		System.out.println(thirdMax(nums3));
	}

	public int thirdMax(int[] nums) {
		int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
		boolean minPresent = false;
		for (int n : nums) {
			if (n == Integer.MIN_VALUE) {
				minPresent = true;
			}

			if (n > max1) {
				max3 = max2;
				max2 = max1;
				max1 = n;
			} else if (n > max2 && n < max1) {
				max3 = max2;
				max2 = n;
			} else if (n > max3 && n < max2) {
				max3 = n;
			}

		}
		if (max3 != Integer.MIN_VALUE) {
			return max3;
		} else {
			if (minPresent && max2 != Integer.MIN_VALUE) {
				return max3;
			} else {
				return max1;
			}
		}
	}
}