[leetcode] 383. Ransom Note

Description

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

分析

题目的意思是： 给定连个random字符串和一个magazine字符串，问random可不可以由magazine字符串组成。

• 如果理解题意了，用一个map就可以搞定了，详情请看代码。

C++实现

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        unordered_map<char, int> m;
        for(auto c:magazine){
            m[c]++;
        }
        for(auto c:ransomNote){
            if(--m[c]<0) return false;
        }
        return true;
    }
};

Python实现

思路就是用hash map实现，一个一个的比对。

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        m = defaultdict(int)
        for ch in magazine:
            m[ch]+=1
        for ch in ransomNote:
            m[ch]-=1
            if m[ch]<0:
                return False
        return True

参考文献

[LeetCode] Ransom Note 赎金条