leetCode 383. Ransom Note 字符串
2016-08-11 14:06
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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. 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.
有一个随机串,有一个大串。判断随机串是否为大串的组成部分。
随机串某一字符的个数必须小于大串。随机串中出现的字符大串中必须都有。
思路:
用map/unordered_map来处理大串,将字符的个数以及种类记录在map/unordered_map中。然后进行判断。
代码如下:
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题目大意:
有一个随机串,有一个大串。判断随机串是否为大串的组成部分。
随机串某一字符的个数必须小于大串。随机串中出现的字符大串中必须都有。
思路:
用map/unordered_map来处理大串,将字符的个数以及种类记录在map/unordered_map中。然后进行判断。
代码如下:
class Solution { public: bool canConstruct(string ransomNote, string magazine) { if(ransomNote.size() == 0) return true; unordered_map<char,int> m; for(int i = 0;i < magazine.size();i++) { m[magazine[i]]++; } for(int i = 0 ; i < ransomNote.size(); i++) { if(m.find(ransomNote[i]) == m.end() || m[ransomNote[i]] == 0 ) return false; m[ransomNote[i]]--; } return true; } };经过测试126组数据,使用map耗时132ms,使用unordered_map耗时84ms。所以在不需要map有序的情况下,使用unordered_map是首选。
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