leetcode-Array篇easy难度之字符串组合分析

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 resource = new HashMap<>(32);
        for(char eachChar: chars.toCharArray()){
            resource.put( eachChar, resource.getOrDefault(eachChar,0)+1);
        }
        for(String eachWord: words){
            Map 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 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