383. Ransom Note

最新推荐文章于 2025-05-09 23:49:38 发布

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本文介绍了一种方法来判断是否能从给定的杂志字符串中构造出勒索信字符串。通过使用不同的C++实现方式，包括利用向量进行字符计数比较、查找并移除字符等方法。

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

vector solution

把每个字符转化成一个下表！

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> m(128,0);
        for(char a : magazine)
        m[a]++;                           //a隐式的转换成int
        for(char b : ransomNote)
        {
            if(m[b] == 0)
            return false;
            else
            m[b]--;
        }
        return true;
    }
};

另一种找下表的方式

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> m(26,0);
        vector<int> n(26,0);
        for(char a : magazine)
        m[a - 'a']++;
        for(char b : ransomNote)
        n[b - 'a']++;
        for(int i = 0; i < 26; i++)
        if(m[i] < n[i])
        return false;
        
        return true;
        
    }
};

？？？ C++的一些东西还不熟

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        for(int i=0; i<ransomNote.size(); i++)
        {
            std::size_t found = magazine.find_first_of(ransomNote[i]);
            if(found==std::string::npos)
                return false;
            else
            {
                magazine.erase(magazine.begin()+found);
            }
        }
        return true;
    }
};

更好理解一些的~

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        for (auto c: ransomNote) {
            auto it = std::find(magazine.begin(), magazine.end(), c);  //find 的用法！
            if (it != magazine.end()) {
                magazine.erase(it);
            } else {
                return false;
            }
        }
        return true;
    }
};