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.
func thirdMax(_ nums: [Int]) -> Int {
var nums = nums.sorted()
var tmp = 2
var max = 0
for i in (0..<nums.count).reversed() {
if i-1 >= 0 && nums[i-1] < nums[i] {
tmp -= 1
if tmp == 0 {
max = nums[i-1]
}
}
}
return tmp <= 0 ? max : nums[nums.count-1]
}
