Leetcode刷题记录_20181024

169.Majority Element

建立字典，统计数字数目，满足要求的输出。

 1 class Solution(object):
 2     def majorityElement(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 7         n = len(nums)
 8         if n == 0:return None
 9         dict1 = {}
10         for i in range(len(nums)):
11             if nums[i] in dict1:
12                 dict1[nums[i]] +=1
13             else:
14                 dict1[nums[i]] = 1
15         for i,j in dict1.items():
16             if j > n/2:
17                 return i
18         return None
View Code

还有一种快速方法，先排序，返回nums[n/2]

171.Excel Sheet Column Number

Excel列名与数字对应关系，反向计算即可；

 1 class Solution(object):
 2     def titleToNumber(self, s):
 3         """
 4         :type s: str
 5         :rtype: int
 6         """
 7         flag = 0
 8         result = 0
 9         for i in s[::-1]:
10             temp = ord(i)-64
11             result = result + temp*26**flag
12             flag +=1
13         return result
View Code

172. Factorial Trailing Zeroes

关键在于数列中存在多少个5就等于多少个0

0的个数等于 n/5+n/5/5+n/5/5/5...直到n/5等于0

 1 class Solution(object):
 2     def trailingZeroes(self, n):
 3         """
 4         :type n: int
 5         :rtype: int
 6         """
 7         result = 0
 8         while n>0:
 9             n = n/5
10             result +=n
11         return result
View Code

189.Rotate Array 

旋转列表n步

可用pop和insert组合

可用切片，然后相加赋值

 1 class Solution(object):
 2     def rotate(self, nums, k):
 3         """
 4         :type nums: List[int]
 5         :type k: int
 6         :rtype: void Do not return anything, modify nums in-place instead.
 7         """
 8         """
 9         k = k%len(nums)
10         for i in range(k):
11             temp = nums.pop()
12             nums.insert(0,temp)
13         """
14         k = k%len(nums)
15         nums[:] = nums[-k:] + nums[:-k]
View Code

 

posted @ 2018-10-24 21:00  adminyzz  阅读(70)  评论(0)  编辑  收藏  举报