414. Third Maximum Number

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?

解題

Runtime: 5 ms, faster than 77.45%

Memory Usage: 2.9 MB, less than 72.06%

func thirdMax(nums []int) int {
    max1 := math.MinInt64
    max2 := math.MinInt64
    max3 := math.MinInt64
    
    for _, val := range nums {
        if val > max1  {
            max3 = max2
            max2 = max1
            max1 = val
        } else if val > max2 && val!= max1 {
            max3 = max2 
            max2 = val
        } else if val > max3 && val!=max2 && val!=max1 {
            max3 = val
        }
    }
    
    if max3==math.MinInt64 { return max1 }
    
    return max3
}