LeetCode--383--赎金信
问题描述:
给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串,判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成。如果可以构成,返回 true ;否则返回 false。
(题目说明:为了不暴露赎金信字迹,要从杂志上搜索各个需要的字母,组成单词来表达意思。)
注意:
你可以假设两个字符串均只含有小写字母。
canConstruct("a", "b") -> false canConstruct("aa", "ab") -> false canConstruct("aa", "aab") -> true
方法:
1 from collections import Counter 2 class Solution(object): 3 def canConstruct(self, ransomNote, magazine): 4 """ 5 :type ransomNote: str 6 :type magazine: str 7 :rtype: bool 8 """ 9 dic1 = Counter(ransomNote) 10 dic2 = Counter(magazine) 11 #dic1,dic2 = map(Counter,(ransomNote,magazine) 12 for key in dic1.keys(): 13 if key in dic2 and dic2[key] <dic1[key] or key not in dic2: 14 return False 15 return True
官方:set(ransomNote) 建立dic存放ransomNote词频计数,用magazine.count(val) 和 dic中的val做对比
1 class Solution(object): 2 def canConstruct(self, ransomNote, magazine): 3 """ 4 :type ransomNote: str 5 :type magazine: str 6 :rtype: bool 7 """ 8 x=set(i for i in ransomNote) 9 dic={} 10 for i in x: 11 dic[i]=ransomNote.count(i) 12 for k,v in dic.items(): 13 if magazine.count(k)<v: 14 return False 15 return True
2018-09-28 15:55:36