LeetCode：383. Ransom Note

051104

题目

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

我的解题思路

这道题是查看字符串magazines是否包含了ransom note string，不管其出现的顺序只要每个字符及其次数能组会起来就行。因此，我们可以对magazines求一个map，存储其字符及其该字符出现的次数。之后再对ransom note string的字符进行判断，当ransom note string中出现该字符则减掉map中的数，若最后所有的字符对应的次数都大于等于0，则返回true，否则false。

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        unordered_map<char, int> map(26);
        for (int i = 0; i < magazine.size(); ++i)
            ++map[magazine[i]];
        for (int j = 0; j < ransomNote.size(); ++j)
            if (--map[ransomNote[j]] < 0)
                return false;
        return true;
    }
};

去看了大佬的解答，不用map之类的stl库实现，而是只用了string的函数。tql~

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        for(char c: ransomNote){
            int found = magazine.find(c);
            if(found == string::npos){
                return false;
            }
            magazine[found] = ' ';  
        }
        return true;
    }
};