Merge pull request SharingSource#738 from SharingSource/ac_oier

✨feat: add 1684

Author: SharingSource

Index/模拟.md

@@ -191,6 +191,7 @@
 | [1662. 检查两个字符串数组是否相等](https://leetcode.cn/problems/check-if-two-string-arrays-are-equivalent/) | [LeetCode 题解链接](https://leetcode.cn/problems/check-if-two-string-arrays-are-equivalent/solution/by-ac_oier-h0l6/) | 简单 | 🤩🤩🤩🤩 |
 | [1672. 最富有客户的资产总量](https://leetcode-cn.com/problems/richest-customer-wealth/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/richest-customer-wealth/solution/by-ac_oier-ai19/) | 简单 | 🤩🤩🤩🤩 |
 | [1678. 设计 Goal 解析器](https://leetcode.cn/problems/goal-parser-interpretation/) | [LeetCode 题解链接](https://leetcode.cn/problems/goal-parser-interpretation/solution/by-ac_oier-a00y/) | 简单 | 🤩🤩🤩🤩 |
+| [1684. 统计一致字符串的数目](https://leetcode.cn/problems/count-the-number-of-consistent-strings/) | [LeetCode 题解链接](https://leetcode.cn/problems/count-the-number-of-consistent-strings/solution/by-ac_oier-j2kj/) | 简单 | 🤩🤩🤩🤩 |
 | [1688. 比赛中的配对次数](https://leetcode-cn.com/problems/count-of-matches-in-tournament/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/count-of-matches-in-tournament/solution/gong-shui-san-xie-jian-dan-nao-jin-ji-zh-cx7a/) | 简单 | 🤩🤩🤩🤩🤩 |
 | [1694. 重新格式化电话号码](https://leetcode.cn/problems/reformat-phone-number/) | [LeetCode 题解链接](https://leetcode.cn/problems/reformat-phone-number/solution/by-ac_oier-82xq/) | 简单 | 🤩🤩🤩🤩 |
 | [1700. 无法吃午餐的学生数量](https://leetcode.cn/problems/number-of-students-unable-to-eat-lunch/) | [LeetCode 题解链接](https://leetcode.cn/problems/number-of-students-unable-to-eat-lunch/solution/by-ac_oier-rvc3/) | 简单 | 🤩🤩🤩🤩🤩 |

LeetCode/1681-1690/1684. 统计一致字符串的数目（简单）.md

@@ -0,0 +1,160 @@
+### 题目描述
+
+这是 LeetCode 上的 **[1684. 统计一致字符串的数目](https://leetcode.cn/problems/count-the-number-of-consistent-strings/solution/by-ac_oier-j2kj/)** ，难度为 **简单**。
+
+Tag : 「模拟」、「位运算」
+
+
+
+给你一个由不同字符组成的字符串 `allowed` 和一个字符串数组 `words`。如果一个字符串的每一个字符都在 `allowed` 中，就称这个字符串是 一致字符串 。
+
+请你返回 `words` 数组中 一致字符串 的数目。
+
+示例 1：
+```
+输入：allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
+
+输出：2
+
+解释：字符串 "aaab" 和 "baa" 都是一致字符串，因为它们只包含字符 'a' 和 'b' 。
+```
+示例 2：
+```
+输入：allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
+
+输出：7
+
+解释：所有字符串都是一致的。
+```
+示例 3：
+```
+输入：allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
+
+输出：4
+
+解释：字符串 "cc"，"acd"，"ac" 和 "d" 是一致字符串。
+```
+
+提示：
+* $1 <= words.length <= 10^4$
+* $1 <= allowed.length <= 26$
+* $1 <= words[i].length <= 10$
+* `allowed` 中的字符 互不相同 。
+* `words[i]` 和 `allowed` 只包含小写英文字母。
+
+---
+
+### 模拟
+
+根据题意模拟即可：为了快速判断某个字符是否在 `allowed` 出现过，我们可以使用 `Set` 结构、定长数组或是一个 `int` 变量（搭配位运算）来对 `allowed` 中出现的字符进行转存。
+
+随后遍历所有的 $words[i]$，统计符合要求的字符串个数。
+
+Java 代码：
+```Java
+class Solution {
+    public int countConsistentStrings(String allowed, String[] words) {
+        boolean[] hash = new boolean[26];
+        for (char c : allowed.toCharArray()) hash[c - 'a'] = true;
+        int ans = 0;
+        out:for (String s : words) {
+            for (char c : s.toCharArray()) {
+                if (!hash[c - 'a']) continue out;
+            }
+            ans++;
+        }
+        return ans;
+    }
+}
+```
+Java 代码：
+```Java
+class Solution {
+    public int countConsistentStrings(String allowed, String[] words) {
+        int hash = 0, ans = 0;
+        for (char c : allowed.toCharArray()) hash |= (1 << (c - 'a'));
+        out:for (String s : words) {
+            for (char c : s.toCharArray()) {
+                if (((hash >> (c - 'a')) & 1) == 0) continue out;
+            }
+            ans++;
+        }
+        return ans;
+    }
+}
+```
+TypeScript 代码：
+```TypeScript
+function countConsistentStrings(allowed: string, words: string[]): number {
+    const sset = new Set<string>()
+    for (const c of allowed) sset.add(c)
+    let ans = 0
+    out:for (const s of words) {
+        for (const c of s) {
+            if (!sset.has(c)) continue out
+        }
+        ans++
+    }
+    return ans
+}
+```
+TypeScript 代码：
+```TypeScript
+function countConsistentStrings(allowed: string, words: string[]): number {
+    let hash = 0, ans = 0
+    for (const c of allowed) hash |= (1 << (c.charCodeAt(0) - 'a'.charCodeAt(0)))
+    out:for (const s of words) {
+        for (const c of s) {
+            if (((hash >> (c.charCodeAt(0) - 'a'.charCodeAt(0))) & 1) == 0) continue out
+        }
+        ans++
+    }
+    return ans
+}
+```
+Python3 代码：
+```Python3
+class Solution:
+    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
+        sset = set([c for c in allowed])
+        ans = 0
+        for s in words:
+            ok = True
+            for c in s:
+                if c not in sset:
+                    ok = False
+                    break
+            ans += 1 if ok else 0
+        return ans
+```
+Python3 代码：
+```Python3
+class Solution:
+    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
+        num, ans = 0, 0
+        for c in allowed:
+            num |= (1 << (ord(c) - ord('a')))
+        for s in words:
+            ok = True
+            for c in s:
+                if not (num >> (ord(c) - ord('a')) & 1):
+                    ok = False
+                    break
+            ans += 1 if ok else 0
+        return ans
+```
+* 时间复杂度：$O(m + \sum_{i = 0}^{n - 1}words[i].length)$，其中 $m$ 为 `allowed` 长度，$n$ 为 `words` 长度
+* 空间复杂度：$O(m)$
+
+---
+
+### 最后
+
+这是我们「刷穿 LeetCode」系列文章的第 `No.1684` 篇，系列开始于 2021/01/01，截止于起始日 LeetCode 上共有 1916 道题目，部分是有锁题，我们将先把所有不带锁的题目刷完。
+
+在这个系列文章里面，除了讲解解题思路以外，还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。
+
+为了方便各位同学能够电脑上进行调试和提交代码，我建立了相关的仓库：https://github.com/SharingSource/LogicStack-LeetCode 。
+
+在仓库地址里，你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。
+