[LeetCode]383. Ransom Note(赎金票据)
2017-04-07 00:14
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383. 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.(给定一个任意的赎金票据字符串和另一个包含杂志上字母字符串,写一个函数,如果可以从杂志中构建赎金票据,则返回true; 否则会返回false。)
Each letter in the magazine string can only be used once in your ransom note.
(杂志字符串中的每个字母只能在您的赎金票据中使用一次。)
Note:
You may assume that both strings contain only lowercase letters.
(你可以假定两个字符串只包含小写字母。)
canConstruct("a", "b") -> false canConstruct("aa", "ab") -> false canConstruct("aa", "aab") -> true
题会做,但是题目刚开始没看懂,看了一个博客理解才题目的意思传送门:
题目叫做Ransom Note,勒索信,刚开始我还没理解这个题目的意思,尤其这个标题,和magazine有啥关系呢?后来仔细想想,才慢慢理解。勒索信,为了不暴露字迹,就从杂志上搜索各个需要的字母,组成单词来表达的意思。这样来说,题目也就清晰了,判断杂志上的字是否能够组成勒索信需要的那些字符。
代码1:
C++ #include <iostream> #include <string> #include <vector> using namespace std; class Solution { public: bool canConstruct(string ransomNote, string magazine) { if (ransomNote.length() > magazine.length()) return false; //考虑了大小写字母,如果只考虑小写字母把128改成26 下面验证的每个字符-97就好 vector <int> res(128, 0);//A-Z 65-90 a-z 97-122 for(char m : magazine) res[m]++; for(char r : ransomNote){ if(res[r] == 0) return false; res[r]--; } return true; } }; int main() { Solution a; string ransomNote,magazine; cin >> ransomNote >> magazine; cout << a.canConstruct(ransomNote, magazine) << endl; return 0; }
代码2:·
bool canConstruct(String ransomNote, String magazine) { int[] arr = new int[26]; for (int i = 0; i < magazine.length(); i++) { arr[magazine.charAt(i) - 'a']++; } for (int i = 0; i < ransomNote.length(); i++) { if(--arr[ransomNote.charAt(i)-'a'] < 0) { return false; } } return true; }
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