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：给两个string，判断第一个string能否由第二个string里面所含有的字母组成，
第二个string里面的所有字母只能使用一次
分析：建立一个hash数组，对第二个string遍历并记录每个字符出现的次数，然后遍历第一个string，
如果有出现hash里面不存在的字符，那么return false
1 class Solution { 2 public: 3 bool canConstruct(string ransomNote, string magazine) { 4 int a[26] = {0}, lenr = ransomNote.length(), lenm = magazine.length(), i; 5 for(i = 0; i < lenm; i++) 6 a[magazine[i] - 'a']++; 7 for(i = 0; i < lenr; i++) 8 if(a[ransomNote[i] - 'a'] == 0) 9 return false; 10 else 11 a[ransomNote[i] - 'a']--; 12 return true; 13 } 14 };