Add solution 417、916

leetcode/0916.Word-Subsets/README.md

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+# [916. Word Subsets](https://leetcode.com/problems/word-subsets/)
+
+
+## 题目
+
+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. `1 <= A.length, B.length <= 10000`
+2. `1 <= A[i].length, B[i].length <= 10`
+3. `A[i]` and `B[i]` consist only of lowercase letters.
+4. All words in `A[i]` are unique: there isn't `i != j` with `A[i] == A[j]`.
+
+## 题目大意
+
+我们给出两个单词数组 A 和 B。每个单词都是一串小写字母。现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称单词 b 是单词 a 的子集。 例如，“wrr” 是 “warrior” 的子集，但不是 “world” 的子集。如果对 B 中的每一个单词 b，b 都是 a 的子集，那么我们称 A 中的单词 a 是通用的。你可以按任意顺序以列表形式返回 A 中所有的通用单词。
+
+## 解题思路
+
+- 简单题。先统计出 B 数组中单词每个字母的频次，再在 A 数组中依次判断每个单词是否超过了这个频次，如果超过了即输出。
+
+## 代码
+
+```go
+package leetcode
+
+func wordSubsets(A []string, B []string) []string {
+	var counter [26]int
+	for _, b := range B {
+		var m [26]int
+		for _, c := range b {
+			j := c - 'a'
+			m[j]++
+		}
+		for i := 0; i < 26; i++ {
+			if m[i] > counter[i] {
+				counter[i] = m[i]
+			}
+		}
+	}
+	var res []string
+	for _, a := range A {
+		var m [26]int
+		for _, c := range a {
+			j := c - 'a'
+			m[j]++
+		}
+		ok := true
+		for i := 0; i < 26; i++ {
+			if m[i] < counter[i] {
+				ok = false
+				break
+			}
+		}
+		if ok {
+			res = append(res, a)
+		}
+	}
+	return res
+}
+```