Add solution 0609、0692、0890、1048、1442、1738

leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern.go

@@ -0,0 +1,32 @@
+package leetcode
+
+func findAndReplacePattern(words []string, pattern string) []string {
+	res := make([]string, 0)
+	for _, word := range words {
+		if match(word, pattern) {
+			res = append(res, word)
+		}
+	}
+	return res
+}
+
+func match(w, p string) bool {
+	if len(w) != len(p) {
+		return false
+	}
+	m, used := make(map[uint8]uint8), make(map[uint8]bool)
+	for i := 0; i < len(w); i++ {
+		if v, ok := m[p[i]]; ok {
+			if w[i] != v {
+				return false
+			}
+		} else {
+			if used[w[i]] {
+				return false
+			}
+			m[p[i]] = w[i]
+			used[w[i]] = true
+		}
+	}
+	return true
+}

leetcode/0890.Find-and-Replace-Pattern/890. Find and Replace Pattern_test.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question890 struct {
+	para890
+	ans890
+}
+
+// para 是参数
+// one 代表第一个参数
+type para890 struct {
+	words   []string
+	pattern string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans890 struct {
+	one []string
+}
+
+func Test_Problem890(t *testing.T) {
+
+	qs := []question890{
+
+		{
+			para890{[]string{"abc", "deq", "mee", "aqq", "dkd", "ccc"}, "abb"},
+			ans890{[]string{"mee", "aqq"}},
+		},
+
+		{
+			para890{[]string{"a", "b", "c"}, "a"},
+			ans890{[]string{"a", "b", "c"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 890------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans890, q.para890
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findAndReplacePattern(p.words, p.pattern))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0890.Find-and-Replace-Pattern/README.md

@@ -0,0 +1,78 @@
+# [890. Find and Replace Pattern](https://leetcode.com/problems/find-and-replace-pattern/)
+
+
+## 题目
+
+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**.
+
+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.
+
+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.
+
+**Example 1:**
+
+```
+Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
+Output: ["mee","aqq"]
+Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
+"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
+```
+
+**Example 2:**
+
+```
+Input: words = ["a","b","c"], pattern = "a"
+Output: ["a","b","c"]
+```
+
+**Constraints:**
+
+- `1 <= pattern.length <= 20`
+- `1 <= words.length <= 50`
+- `words[i].length == pattern.length`
+- `pattern` and `words[i]` are lowercase English letters.
+
+## 题目大意
+
+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。
+
+## 解题思路
+
+- 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。
+
+## 代码
+
+```go
+package leetcode
+
+func findAndReplacePattern(words []string, pattern string) []string {
+	res := make([]string, 0)
+	for _, word := range words {
+		if match(word, pattern) {
+			res = append(res, word)
+		}
+	}
+	return res
+}
+
+func match(w, p string) bool {
+	if len(w) != len(p) {
+		return false
+	}
+	m, used := make(map[uint8]uint8), make(map[uint8]bool)
+	for i := 0; i < len(w); i++ {
+		if v, ok := m[p[i]]; ok {
+			if w[i] != v {
+				return false
+			}
+		} else {
+			if used[w[i]] {
+				return false
+			}
+			m[p[i]] = w[i]
+			used[w[i]] = true
+		}
+	}
+	return true
+}
+```