【LeetCode】 1160. Find Words That Can Be Formed by Characters 拼写单词（Easy）（JAVA）

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

题目描述：

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

题目大意

给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。

假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。

注意：每次拼写时，chars 中的每个字母都只能用一次。

返回词汇表 words 中你掌握的所有单词的 长度之和。

解题方法

1、遍历记录下 chars 里面每个字母的个数
2、遍历每个单词的每个字母，确认是否在 chars 字母个数的范围内

class Solution {
    public int countCharacters(String[] words, String chars) {
        int[] count = new int[26];
        for (int i = 0; i < chars.length(); i++) {
            count[chars.charAt(i) - 'a']++;
        }
        int res = 0;
        for (int i = 0; i < words.length; i++) {
            int[] temp = count.clone();
            int j = 0;
            for (; j < words[i].length(); j++) {
                int cur = words[i].charAt(j) - 'a';
                temp[cur]--;
                if (temp[cur] < 0) break;
            }
            if (j == words[i].length()) res += words[i].length();
        }
        return res;
    }
}

执行用时 : 9 ms, 在所有 Java 提交中击败了 72.91% 的用户
内存消耗 : 42.1 MB, 在所有 Java 提交中击败了 5.08% 的用户