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78 lines
2.7 KiB
Markdown
78 lines
2.7 KiB
Markdown
# [890. Find and Replace Pattern](https://leetcode.com/problems/find-and-replace-pattern/)
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## 题目
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Given a list of strings `words` and a string `pattern`, return *a list of* `words[i]` *that match* `pattern`. You may return the answer in **any order**.
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A word matches the pattern if there exists a permutation of letters `p` so that after replacing every letter `x` in the pattern with `p(x)`, we get the desired word.
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Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.
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**Example 1:**
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```
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Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
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Output: ["mee","aqq"]
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Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
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"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
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```
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**Example 2:**
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```
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Input: words = ["a","b","c"], pattern = "a"
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Output: ["a","b","c"]
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```
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**Constraints:**
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- `1 <= pattern.length <= 20`
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- `1 <= words.length <= 50`
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- `words[i].length == pattern.length`
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- `pattern` and `words[i]` are lowercase English letters.
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## 题目大意
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你有一个单词列表 words 和一个模式 pattern，你想知道 words 中的哪些单词与模式匹配。如果存在字母的排列 p ，使得将模式中的每个字母 x 替换为 p(x) 之后，我们就得到了所需的单词，那么单词与模式是匹配的。（回想一下，字母的排列是从字母到字母的双射：每个字母映射到另一个字母，没有两个字母映射到同一个字母。）返回 words 中与给定模式匹配的单词列表。你可以按任何顺序返回答案。
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## 解题思路
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- 按照题目要求，分别映射两个字符串，words 字符串数组中的字符串与 pattern 字符串每个字母做映射。这里用 map 存储。题目还要求不存在 2 个字母映射到同一个字母的情况，所以再增加一个 map，用来判断当前字母是否已经被映射过了。以上 2 个条件都满足即代表模式匹配上了。最终将所有满足模式匹配的字符串输出即可。
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## 代码
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```go
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package leetcode
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func findAndReplacePattern(words []string, pattern string) []string {
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res := make([]string, 0)
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for _, word := range words {
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if match(word, pattern) {
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res = append(res, word)
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}
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}
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return res
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}
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func match(w, p string) bool {
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if len(w) != len(p) {
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return false
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}
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m, used := make(map[uint8]uint8), make(map[uint8]bool)
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for i := 0; i < len(w); i++ {
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if v, ok := m[p[i]]; ok {
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if w[i] != v {
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return false
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}
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} else {
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if used[w[i]] {
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return false
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}
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m[p[i]] = w[i]
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used[w[i]] = true
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}
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}
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return true
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}
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``` |