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 Subscribe to see which companies asked this question
解题思路： 给定一个随机字符串，另外一个字符串是否包含这个字符串所有元素个数。建一个26个字母的数组，对随机字符串进行求统计，然后再对另外一个字符串求统计，如果没有出现负值表示true，出现负值表示false。
class Solution { public: bool canConstruct(string ransomNote, string magazine) { vector<int> charcnt(26,0); for(int i = 0; i < ransomNote.size(); ++i) { charcnt[ransomNote[i]-'a'] --; } for(int i = 0; i < magazine.size(); ++i) { charcnt[magazine[i]-'a'] ++; } for(int i = 0; i < 26; ++i) { if(charcnt[i] < 0) { return false; } } return true; } };