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thirdMax.py
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thirdMax.py
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# https://leetcode.com/problems/third-maximum-number/description/
# 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.
from collections import Counter as c
def thirdMax(nums):
"""
:type nums: List[int]
:rtype: int
"""
d = c(nums)
key = d.keys()
n = len(key)
if n < 3:
return max(key)
maxi1, maxi2, maxi3 = key[0], float('-inf'), float('-inf')
for num in key[1:]:
if num > maxi1:
maxi3 = maxi2
maxi2 = maxi1
maxi1 = num
elif num > maxi2:
maxi3 = maxi2
maxi2 = num
elif num > maxi3:
maxi3 = num
return maxi3
assert thirdMax([-3, -2, -1]) == -3
assert thirdMax([3, 2, 1]) == 1
assert thirdMax([1, 2]) == 2
assert thirdMax([2, 2, 3, 1]) == 1