1160. Find Words That Can Be Formed by Characters

问题：

给定一个字符串数组，和一个字典字符串，

若字符串数组中的一个字符串中的所有字符，能够在字典字符串中一一对应找到（字典中每个字符只能用一次），则返回这个字符串的长度。

最终返回满足条件的所有字符串长度之和。

Example 1:
Input: words = ["cat","bt","hat","tree"], chars = "atach"
Output: 6
Explanation: 
The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.

Example 2:
Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
Output: 10
Explanation: 
The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10. 

Note:
1 <= words.length <= 1000
1 <= words[i].length, chars.length <= 100
All strings contain lowercase English letters only.

　　

解法：

用dict[26]保存字典字符串中各个字符出现的次数。

再轮询字符串数组的每个字符串，

在dict中查找对应的字符，找到一个dict[i]--

因此，dict每次对比的时候需要copy一份dic用来临时判断，以不影响下一个字符串的判断。

代码参考：

class Solution {
public:
    int getcount(string str, vector<int> dict){
        int cnt=0;
        vector<int> dic(dict);
        for(char s:str){
            if(dic[s-'a']<=0) return 0;
            dic[s-'a']--;
            cnt++;
        }
        return cnt;
    }
    int countCharacters(vector<string>& words, string chars) {
        vector<int> dict(26);
        int res=0;
        for(char c:chars){
            dict[c-'a']++;
        }
        for(string str:words){
            res+=getcount(str, dict);
        }
        return res;
    }
};