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:

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Input: [3, 2, 1]
Output: 1
Explanation: The third maximum is 1.

Example 2:

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Input: [1, 2]
Output: 2
Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

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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.

思路

粗暴地进行三次求最大值操作，把第三大的数找出来。这堆代码写得很乱，详见这个简洁的

其实一趟操作完全可以做到。详见这个解法

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class Solution(object):
    def thirdMax(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        maxnum = []
        for i in range(3):
            while (len(nums)):
                tmp = max(nums)
                nums.remove(tmp)
                if (len(nums)):
                    if (max(nums) != tmp):
                        break
                else:
                    break
            
            maxnum.append(tmp)
            
            print nums
            if (not len(nums) and i != 2):
                return maxnum[0]
        return maxnum[-1]