随笔档案「2019年9月」 - Sempron2800+

leetcode1190

摘要：参考：https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/discuss/382775/Python3-straightforward-and-easiest-on-discussion 阅读全文

posted @ 2019-09-15 22:45 Sempron2800+ 阅读(322) 评论(0) 推荐(0)

leetcode1189

摘要：算法思想：贪心+Hash。记录b a l o n出现的次数，计算可以构成的单词的最少字符。阅读全文

posted @ 2019-09-15 21:36 Sempron2800+ 阅读(216) 评论(0) 推荐(0)

校招真题练习035 最少立方数之和(小米)

摘要：最少立方数之和题目描述给出一个数字N（0<N<1000000）,将N写成立方数和的形式，求出需要的最少立方数个数。例如N=17，1+8+8 = 17，最少需要3个立方数，则输出3。N= 28,1+1+1+1+8+8+8=28, 需要7个立方数，1+27=28,需要2个立方数，所以最少立方数为2，则阅读全文

posted @ 2019-09-11 09:11 Sempron2800+ 阅读(386) 评论(0) 推荐(0)

leetcode1185

摘要：1 from datetime import datetime 2 3 class Solution: 4 def dayOfTheWeek(self, day: int, month: int, year: int) -> str: 5 days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday","Sund 阅读全文

posted @ 2019-09-11 07:19 Sempron2800+ 阅读(229) 评论(0) 推荐(0)

leetcode1184

摘要：1 class Solution: 2 def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int: 3 if start == destination: 4 return 0 5 elif start < destination: 6 dis1,dis2 = 0,0 7 f 阅读全文

posted @ 2019-09-11 07:08 Sempron2800+ 阅读(170) 评论(0) 推荐(0)

leetcode1175

摘要：1 class Solution: 2 def factorial(self,N): 3 result = 1 4 while N > 0: 5 result *= N 6 N -= 1 7 return result 8 9 def numPrimeArrange... 阅读全文

posted @ 2019-09-10 21:14 Sempron2800+ 阅读(158) 评论(0) 推荐(0)

leetcode1170

摘要：1 import collections 2 import bisect 3 4 class Solution: 5 def getFC(self,word): 6 obj = collections.Counter(word) 7 for i in range(97,123): 8 c = chr(i) 9 if c in obj: 10 return obj[c] 11 12 def numS 阅读全文

posted @ 2019-09-10 16:41 Sempron2800+ 阅读(174) 评论(0) 推荐(0)

leetcode1154

摘要：1 class Solution: 2 def isLeap(self,year): 3 if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0): 4 return True 5 return False 6 7 def dayOfYea... 阅读全文

posted @ 2019-09-10 16:13 Sempron2800+ 阅读(161) 评论(0) 推荐(0)