题解 | 字符串构造判定

from pickle import FALSE
#
# 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
#
class Solution:
    def canConstruct(self , ransomNote: str, magazine: str) -> bool:
        # write code here
        ls = [[0,0] for _ in range(26)]
        for i in range(len(magazine)):
            ls[ord(magazine[i]) - ord('a')][1] += 1
        for j in range(len(ransomNote)):
            ls[ord(ransomNote[j]) - ord('a')][1] -= 1
        for k in range(26):
            if ls[k][1] < 0:
                return False
        return True