题解 | 字符串构造判定

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param ransomNote string字符串 
# @param magazine string字符串 
# @return bool布尔型
#
from collections import Counter
class Solution:
    def canConstruct(self , ransomNote: str, magazine: str) -> bool:
        # write code here
        s = Counter(ransomNote)
        t = Counter(magazine)
        for c in s.keys():#依次查看s中的元素是否都在t中，且数量正确
            if c not in t or s[c]>t[c]:
                return False
        return True