leetcode Ransom Note

leetcode 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.
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canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

题目地址：leetcode Ransom Note

题目大意：给定两个字符串magazines和ransomNote，问是否可以从magazines中抽取字母（每个字母只能用一次）组成ransomNote

思路：只要判断ransomNote字符是不是全部在magazines中即可，用hash。。水

C++

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> cnt(26,0);
        for(char x:magazine) cnt[x-97]++;
        for(char x:ransomNote) if(--cnt[x-97] < 0) return false;
        return true;
    }
};

Java

public class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] cnt = new int[26];
        for (int i = 0; i < magazine.length(); i++) cnt[magazine.charAt(i) - 97]++;
for (int i = 0; i < ransomNote.length(); i++) if (--cnt[ransomNote.charAt(i) - 97] < 0) return false;
return true;
    }
}

Python

class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        cnt = collections.Counter(magazine)
        for letter in ransomNote:
            cnt[letter] -= 1
            if cnt[letter] <0: return False
        return True

本题是leetcode 383 Ransom Note 的题解

更多题解可以查看：https://www.hrwhisper.me/leetcode-algorithm-solution/