LeetCode: 383 Ransom Note(easy) C++

玲珑子_a

于 2018-12-14 11:30:57 发布

阅读量243

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode学习笔记

本文链接：https://blog.youkuaiyun.com/lxin_liu/article/details/85000354

LeetCode学习笔记专栏收录该内容

30 篇文章

订阅专栏

本文探讨了一种解决勒索信问题的字符串匹配算法。该算法检查是否能从杂志字符串中构建出勒索信字符串，确保杂志字符串中的每个字符在勒索信中仅使用一次。通过两个解决方案对比，一种逐个字符查找并移除，另一种预先统计杂志字符串中各字符出现次数，实现高效判断。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

1 canConstruct("a", "b") -> false
2 canConstruct("aa", "ab") -> false
3 canConstruct("aa", "aab") -> true

代码:

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        if (ransomNote == "")
            return true;
        for (auto c : ransomNote){
            int position = magazine.find_first_of(c); 
            if (position != magazine.npos)
                magazine.erase(position, 1);
            else
                return false;        
        }
       return true;    
    }
};

别人的：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int mp[26] = {0};
        for(auto c : magazine) {
            mp[c - 'a']++;
        }
        for(auto r : ransomNote) {
            mp[r - 'a']--;
            if(mp[r - 'a'] < 0) return false;
        }
        return true;
    }
};