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

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        if(ransomNote.length() > magazine.length())
        return false;
        else
        {
        	for(int i=0;i<ransomNote.length();i++)
        	{
        		bool flag = false;
        		for(int j=0;j<magazine.length();j++)
        		{
        			if(ransomNote[i] == magazine[j])
        			{
        				magazine[j] = '0';
        				flag = true;
        				break;
}
}
if(flag == false)
return false;
}

return true;
}
    }
};

以下是来自评论区sharmilas的代码 使用了hashmap 截取下来以作参考

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {

int i=0,j=0;
map <char ,int> h;
while(i<m.length())
{    h[m[i]]++;
i++;
}
i=0;

while(i<r.length())
{
if(h.find(r[i])!=h.end())
{
h[r[i]]=h[r[i]]-1;
if(h[r[i]]==0)
h.erase(h.find(r[i]));
}
else
return false;
i++;
}
return true;
}
};