383. Ransom Note 赎金信

@TOC

[LeetCode]

• Difficulty: Easy

# 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
``````

# 解题方法

# Java解法

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

AC：18 ms

# Python解法

``````class Solution:
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
rcount = collections.Counter(ransomNote)
mcount = collections.Counter(magazine)
for r, c in rcount.items():
if c > mcount[r]:
return False
return True
``````

# 日期

2017 年 1 月 7 日 2018 年 11 月 14 日 —— 很严重的雾霾