### 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


### 解法一：用列表

#### 代码

class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
for (int i = 0; i < magazine.length(); i++) {
}
for (int i = 0; i < ransomNote.length(); i++) {
if (!mgz.remove((Character)ransomNote.charAt(i))) { return false; }
}
return true;
}
}


### 解法二：统计字符频率

#### 代码

class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
int[] freq = countFreq(magazine);
for (int i = 0; i < ransomNote.length(); i++) {
int offset = ransomNote.charAt(i) - 'a';
if (freq[offset] > 0) {
--freq[offset];
} else {
return false;
}
}
return true;
}
private int[] countFreq(String magazine) {
int[] freq = new int[26];
for (int i = 0; i < magazine.length(); i++) {
++freq[magazine.charAt(i)-'a'];
}
return freq;
}
}