383. Ransom Note【easy】

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

解法一：

 1 class Solution {
 2 public:
 3     bool canConstruct(string ransomNote, string magazine) {
 4         map<char
 
, int> m_rans;
 5         
 6         for (int i = 0; i < ransomNote.length(); ++i) {
 7             ++m_rans[ransomNote[i]];
 8         }
 9         
10         for (int i = 0; i < magazine.length(); ++i) {
11             if (m_rans.find(magazine[i]) != m_rans.end()) {
12                 --m_rans[magazine[i]];
 
13             }
14         }
15         
16         for (map<char, int>::iterator it = m_rans.begin(); it != m_rans.end(); ++it) {
17             if (it->second > 0) {
18                 return false;
19             }
20         }
21         
22         return true;
23     }
24 };

解法二：

 1 public class Solution {
 2     public boolean canConstruct(String ransomNote, String magazine) {
 3         int[] arr = new int[26];
 4         for (int i = 0; i < magazine.length(); i++) {
 5             arr[magazine.charAt(i) - ‘a‘]++;
 6         }
 7         for (int i = 0; i < ransomNote.length(); i++) {
 8             if(--arr[ransomNote.charAt(i)-‘a‘] < 0) {
 9                 return false;
10             }
11         }
12         return true;
13     }
14 }

參考@yidongwang 的代碼。

解法三：

 1 class Solution {
 2 public:
 3     bool canConstruct(string ransomNote, string magazine) {
 4         unordered_map<char, int> map(26);
 5         for (int i = 0; i < magazine.size(); ++i)
 6             ++map[magazine[i]];
 7         for (int j = 0; j < ransomNote.size(); ++j)
 8             if (--map[ransomNote[j]] < 0)
 9                 return false;
10         return true;
11     }
12 };

參考@haruhiku 的代碼

解法四：

 1 class Solution {
 2 public:
 3     bool canConstruct(string ransomNote, string magazine) {
 4         vector<int> vec(26, 0);
 5         for (int i = 0; i < magazine.size(); ++i)
 6             ++vec[magazine[i] - ‘a‘];
 7         for (int j = 0; j < ransomNote.size(); ++j)
 8             if (--vec[ransomNote[j] - ‘a‘] < 0)
 9                 return false;
10         return true;
11     }
12 };

參考@haruhiku 的代碼

383. Ransom Note【easy】