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79 lines
2.1 KiB
Markdown
79 lines
2.1 KiB
Markdown
# [1684. Count the Number of Consistent Strings](https://leetcode.com/problems/count-the-number-of-consistent-strings/)
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## 题目
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You are given a string `allowed` consisting of **distinct** characters and an array of strings `words`. A string is **consistent** if all characters in the string appear in the string `allowed`.
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Return *the number of **consistent** strings in the array* `words`.
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**Example 1:**
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```
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Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
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Output: 2
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Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.
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```
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**Example 2:**
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```
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Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
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Output: 7
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Explanation: All strings are consistent.
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```
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**Example 3:**
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```
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Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
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Output: 4
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Explanation: Strings "cc", "acd", "ac", and "d" are consistent.
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```
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**Constraints:**
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- `1 <= words.length <= 104`
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- `1 <= allowed.length <= 26`
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- `1 <= words[i].length <= 10`
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- The characters in `allowed` are **distinct**.
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- `words[i]` and `allowed` contain only lowercase English letters.
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## 题目大意
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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 。
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请你返回 `words` 数组中 一致字符串 的数目。
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## 解题思路
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- 简单题。先将 `allowed` 转化成 map。将 `words` 数组中每个单词的字符都在 map 中查找一遍，如果都存在就累加 res。如果有不存在的字母，不累加。最终输出 res 即可。
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## 代码
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```go
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package leetcode
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func countConsistentStrings(allowed string, words []string) int {
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allowedMap, res, flag := map[rune]int{}, 0, true
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for _, str := range allowed {
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allowedMap[str]++
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}
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for i := 0; i < len(words); i++ {
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flag = true
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for j := 0; j < len(words[i]); j++ {
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if _, ok := allowedMap[rune(words[i][j])]; !ok {
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flag = false
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break
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}
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}
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if flag {
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res++
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}
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}
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return res
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}
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``` |