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

既然是要检查note中的字母是不是全都出现在magazines中,那不妨先遍历magazines一遍,记录下所有字母的出现次数,然后再遍历note,对每个字母检查其是否出现在magazines中,若出现则再检查出现次数是否比magazines中少。

在代码中有几个可以优化的地方:

  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;
    }
};