383. Ransom Note

最新推荐文章于 2025-05-09 23:49:38 发布

FreeeLinux

最新推荐文章于 2025-05-09 23:49:38 发布

阅读量400

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/FreeeLinux/article/details/61423210

Leetcode 专栏收录该内容

64 篇文章

订阅专栏

本文介绍了一种用于判断是否能从给定的杂志字符串中构造出勒索信字符串的算法。通过两次遍历字符串，利用哈希表记录字母出现次数，实现高效匹配。文章包含初始版本与优化后的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

第一次：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        const int len1 = ransomNote.length();
        const int len2 = magazine.length();
        if(len1 == 0 && len2 == 0)
            return true;

        unordered_map<int, int> umap;
        for(int i=0; i<len1; ++i)
            umap[ransomNote[i]]++;
        for(int i=0; i<len2; ++i)
            umap[magazine[i]]--;
        auto it = find_if(umap.begin(), umap.end(), [](const pair<int, int>& pr) 
                                    {return pr.second > 0;} );
        return it == umap.end();
    }
};

优化后：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        const int len1 = ransomNote.length();
        const int len2 = magazine.length();
        if(len1 == 0 && len2 == 0)
            return true;

        unordered_map<int, int> umap;
        for(int i=0; i<len2; ++i)
            umap[magazine[i]]++;
        for(int i=0; i<len1; ++i) {
            if(--umap[ransomNote[i]] < 0)
                return false;
        }
        return true;
    }
};