414. Third Maximum Number
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.
class Solution(object): def thirdMax(self, nums): """ :type nums: List[int] :rtype: int """ a,b,c = -2147483649,-2147483649,-2147483649 for i in range(len(nums)): if nums[i]>a: c = b b = a a = nums[i] elif (nums[i]>b and nums[i]<a): c = b b = nums[i] elif (nums[i]>c and nums[i]<b): c = nums[i] return a if (c==-2147483649 or c==b) else c