## 题目描述

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``````

## 解法一

### 思路

1. 记录magazines中字母出现次数一般用Map，但注意这里magazines中一定是小写字母，所以不妨用一个`int[26]`来存储。判断“字母是否存在”也只需要检查其值是否为`0`即可。
2. 检查出现次数时，对每个note中的字母，只需要将`int[]`中对应位置上的值减一就好了。当遇到`int[]`中对应位置值为0时，只可能是字母在magazines中不存在，或字母在note中出现次数已经超过magazines中出现次数，这两种情况对应都是不满足题目要求的情况，直接返回`False`就好了。

### Python

``````class Solution:
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
has = [0] * 26
for item in magazine:
has[ord(item) - ord('a')] += 1
for item in ransomNote:
if has[ord(item) - ord('a')] == 0:
return False
has[ord(item) - ord('a')] -= 1
return True``````

### Java

``````public class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
int[] has = new int[26];
for (char item : magazine.toCharArray()) {
has[item - 'a']++;
}
for (char item : ransomNote.toCharArray()) {
if (has[item - 'a'] == 0) {
return false;
}
has[item - 'a']--;
}
return true;
}
}``````

### C++

``````class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
vector<int> has(26, 0);
for (char item : magazine) {
has[item - 'a']++;
}
for (char item : ransomNote) {
if (has[item - 'a'] == 0) {
return false;
}
has[item - 'a']--;
}
return true;
}
};``````