LeetCode383赎金信

# coding:utf-8
"""
Name : LeetCode383.py
Author  : qlb
Contect : 17801044486@163.com
Time    : 2021/2/6 9:32
Desc:赎金信
"""
# 解题思路 和四数相加思路类似
# 1分为两部分 将两个字符串分别按照 key(字符) value（字符出现的次数）存储在两个哈希表中
# 2 遍历 保存ransomNote的哈希表 ransomNote每个字符出现的次数小于等于该字符在magazine中出现的次数即可

#Solution1 原始写法
import collections


class Solution1:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        bigDic = {}
        smallDic = {}
        for s1 in magazine:
            if s1 in bigDic:
                bigDic[s1] += 1
            else:
                bigDic[s1] = 1
        for s2 in ransomNote:
            if s2 in smallDic:
                smallDic[s2] += 1
            else:
                smallDic[s2] = 1
        boolCount = 0
        for k,v in smallDic.items():
            if k in bigDic:
                if v <= bigDic[k]:
                    boolCount += 1
        if boolCount == len(smallDic):
            return True
        else:
            return False

# Solution2 效率更高的写法
class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        bigDic = collections.Counter(magazine)
        smallDic = collections.Counter(ransomNote)
        for k,v in smallDic.items():
            if bigDic.get(k,-1) < v:
                return False
        return True