【leetcode】【easy】【每日一题】1160. Find Words That Can Be Formed by Characters

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本文解析了LeetCode上一道关于字符串和字符处理的问题，通过分析如何从给定的字符集中形成有效单词，并计算所有有效单词的总长度。提供了一种使用数组记录字符频率的方法，实现了高效判断和求和。

1160. Find Words That Can Be Formed by Characters

You are given an array of strings words and a string chars.

A string is good if it can be formed by characters from chars (each character can only be used once).

Return the sum of lengths of all good strings in words.

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.

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

思路

利用容器帮助记录。

class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        if(chars.size()<0) return 0;
        int res = 0;
        int charnum[26] = {0};
        for(int i=0; i<chars.size(); ++i){
            ++charnum[chars[i]-'a'];
        }
        for(int i=0; i<words.size(); ++i){
            int wordnum[26] = {0};
            string word = words[i];
            bool isok = true;
            for(int j=0; j<word.size(); ++j){
                int idx = word[j]-'a';
                ++wordnum[idx];
                if(wordnum[idx]>charnum[idx]) {
                    isok = false;
                    break;
                }
            }
            if(isok) res += word.size();
        }
        return res;
    }
};