You are given an array of stringswordsand a stringchars.

A string isgoodifit can be formed bycharacters fromchars(each charactercan only be used once).

Return the sum of lengths of all good strings inwords.

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. 1 <= words.length <= 1000
2. 1 <= words[i].length, chars.length<= 100
3. All strings contain lowercase English letters only.

這道題給了一個單詞陣列 words，還有一個字串 chars，定義了一種好字串，說是能由 chars 中的字母組成的單詞，限定每個字母只能使用一次，不必都使用，求所有好字串的長度之和。既然是 Easy 的身價，自然不會用到太 fancy 的解法，就是一個單純的字母統計問題，建立 chars 字串中每個字母和其出現次數之間的對映，然後遍歷每個單詞，拷貝一個 chars 字串的 HashMap，然後遍歷當前單詞的每個字母，對應字母的對映值減1，若為負數了，表示 chars 中缺少必要的單詞，標記為 false。若最終為 true，則將當前單詞的長度加入結果 res 中即可，參見程式碼如下：

class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        int res = 0;
        unordered_map<char, int> charCnt;
        for (char c : chars) ++charCnt[c];
        for (string word : words) {
            unordered_map<char, int> m = charCnt;
            bool succeed = true;
            for (char c : word) {
                if (--m[c] < 0) {
                    succeed = false;
                    break;
                }
            }
            if (succeed) res += word.size();
        }
        return res;
    }
};
 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1160

參考資料：

https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/

https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/360978/C%2B%2B-Track-Count

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