leetcode 383. Ransom Note

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本文介绍了一种简单的算法，用于判断一个给定的勒索信字符串是否能通过另一个包含杂志上所有字母的字符串构建而成。该算法使用了 map 数据结构来记录杂志字符串中每个字符出现的次数，并逐一检查勒索信中的字符是否能在杂志字符串中找到。

383. Ransom Note

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

简单map

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) 
    {
        map<char, int> mp;
        for (int i = 0; i< magazine.size(); i++)
        {
            mp[magazine[i]] ++;   
        }
        for (int i = 0; i< ransomNote.size(); i++)
        {
            if (mp.find(ransomNote[i]) == mp.end()) return false;
            else if ( --mp[ransomNote[i]] == 0)    mp.erase(ransomNote[i]);
        }
        return true;
    }
};