Easy
Given an integer array nums
, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.
Example 1:
Input: nums = [3,2,1]
Output: 1
Explanation: The first distinct maximum is 3. The second distinct maximum is 2. The third distinct maximum is 1.
Example 2:
Input: nums = [1,2]
Output: 2
Explanation: The first distinct maximum is 2. The second distinct maximum is 1. The third distinct maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: nums = [2,2,3,1]
Output: 1
Explanation: The first distinct maximum is 3. The second distinct maximum is 2 (both 2’s are counted together since they have the same value). The third distinct maximum is 1.
Constraints:
1 <= nums.length <= 104
-231 <= nums[i] <= 231 - 1
Follow up: Can you find an O(n)
solution?
public class Solution {
public int thirdMax(int[] nums) {
long max1 = Long.MIN_VALUE;
long max2 = Long.MIN_VALUE;
long max3 = Long.MIN_VALUE;
for (int i : nums) {
max1 = Math.max(max1, i);
}
for (int i : nums) {
if (i == max1) {
continue;
}
max2 = Math.max(max2, i);
}
for (int i : nums) {
if (i == max1 || i == max2) {
continue;
}
max3 = Math.max(max3, i);
}
return (int) (max3 == Long.MIN_VALUE ? max1 : max3);
}
}