Word Subsets

最新推荐文章于 2023-08-18 11:19:36 发布

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本文介绍了一种高效的算法，用于解决字符串子集匹配问题。通过记录字符串B中每个字母出现的最大次数，筛选出字符串数组A中包含B中所有字符串的元素。文章详细解释了算法的实现过程，并提供了具体示例。

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We are given two arrays A and B of words. Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity. For example, "wrr" is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is universal if for every b in B, b is a subset of a.

Return a list of all universal words in A. You can return the words in any order.

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]

Note:

1 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i] and B[i] consist only of lowercase letters.
All words in A[i] are unique: there isn't i != j with A[i] == A[j].

题目理解：

定义字符串a包含字符串b：对于每一个字母，a中出现的次数大于等于b中出现的次数。给定字符串数组A和B，找出A中所有满足条件的字符串a，a包含B中所有的字符串

解题思路：

如果a包含B中的所有字符串，那么a一定满足最严格的条件，即对于每一个字母，a中出现的次数，大于B中任意一个字符串中出现的次数，换句话说，大于B中这个字母出现次数最多的字符串中出现的次数。

因此字需要记录所有字母，在B中出现的最大次数，满足出现字母数量大于这些最大次数的字符串就是我们要找的

class Solution {
    public List<String> wordSubsets(String[] A, String[] B) {
        int[] record = new int[26];
        for(String str : B){
            int[] local = new int[26];
            for(char ch : str.toCharArray())
                local[ch - 'a']++;
            for(int i = 0; i < 26; i++)
                record[i] = Math.max(record[i], local[i]);
        }
        List<String> res = new ArrayList<>();
        for(String str : A){
            int[] local = new int[26];
            for(char ch : str.toCharArray())
                local[ch - 'a']++;
            boolean flag = true;
            for(int i = 0; i < 26 && flag; i++)
                if(local[i] < record[i])
                    flag = false;
            if(flag)
                res.add(str);
        }
        return res;
    }
}