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Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.
Each letter in the magazine string can only be used once in your ransom note.
Note:
You may assume that both strings contain only lowercase letters.
canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
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勒索信，为了不暴露字迹，就从杂志上搜索各个需要的字母，组成单词来表达的意思。这样来说，题目也就清晰了，判断杂志上的字是否能够组成勒索信需要的那些字符。
这里需要注意的就是杂志上的字符只能被使用一次，不过不用考虑大小写的问题。
有一种最简单的理解就是对于ransomNote里每个字符出现的次数必须小于或者等于该字符在magazine出现的次数。
class Solution(object):def canConstruct(self, ransomNote, magazine):#print ransomNoter = ransomNotem = magazinedic_r = {}dic_m = {}for i in r:dic_r[i] = dic_r.get(i,0) + 1for i in m:dic_m[i] = dic_m.get(i,0) + 1#print dic_r#print dic_mfor k,v in dic_r.items():if k not in dic_m.keys():return Falseif v > dic_m[k]:return Falsereturn True这篇关于383. Ransom Note【E】的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
