关键词

字符串

题目描述

https://leetcode.com/problems/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.

博主提交的代码

class Solution {
    public int countCharacters(String[] words, String chars) {
        int result = 0;
        Map<Character,Integer> resource = new HashMap<>(32);
        for(char eachChar: chars.toCharArray()){
            resource.put( eachChar, resource.getOrDefault(eachChar,0)+1);
        }
        for(String eachWord: words){
            Map<Character,Integer> now = new HashMap<>(32);
            for(char eachChar: eachWord.toCharArray()){
                now.put( eachChar, now.getOrDefault(eachChar,0)+1);
            }
            if(now.size() > resource.size()){
                continue ;
            }
            boolean findIt = true;
            for(Map.Entry<Character,Integer> entry: now.entrySet()){
                if( resource.get(entry.getKey()) == null || resource.get(entry.getKey()) < entry.getValue()){
                    findIt = false;
                    break;
                }
            }
            if(findIt){
                result+= eachWord.length();
            }
            
        }
        return result;
    }
}

其他人优秀的解法

持续关注吧，没有发现太好的