添加内容

website/content/ChapterFour/0290.Word-Pattern.md

@@ -0,0 +1,98 @@
+# [290. Word Pattern](https://leetcode.com/problems/word-pattern/)
+
+## 题目
+
+Given a pattern and a string str, find if str follows the same pattern.
+
+Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.
+
+**Example 1**:
+
+```
+
+Input: pattern = "abba", str = "dog cat cat dog"
+Output: true
+
+```
+
+**Example 2**:
+
+```
+
+Input:pattern = "abba", str = "dog cat cat fish"
+Output: false
+
+```
+
+**Example 3**:
+
+```
+
+Input: pattern = "aaaa", str = "dog cat cat dog"
+Output: false
+
+```
+
+**Example 4**:
+
+```
+
+Input: pattern = "abba", str = "dog dog dog dog"
+Output: false
+
+```
+
+**Note**:
+
+You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
+
+
+## 题目大意
+
+给定一个模式串，判断字符串是否和给定的模式串，是一样的模式。
+
+## 解题思路
+
+这道题用 2 个 map 即可。1 个 map 记录模式与字符串的匹配关系，另外一个 map 记录字符串和模式的匹配关系。为什么需要记录双向的关系呢？因为 Example 4 中，a 对应了 dog，这个时候 b 如果再对应 dog 是错误的，所以这里需要从 dog 查询它是否已经和某个模式匹配过了。所以需要双向的关系。
+
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import "strings"
+
+func wordPattern(pattern string, str string) bool {
+	strList := strings.Split(str, " ")
+	patternByte := []byte(pattern)
+	if pattern == "" || len(patternByte) != len(strList) {
+		return false
+	}
+
+	pMap := map[byte]string{}
+	sMap := map[string]byte{}
+	for index, b := range patternByte {
+		if _, ok := pMap[b]; !ok {
+			if _, ok = sMap[strList[index]]; !ok {
+				pMap[b] = strList[index]
+				sMap[strList[index]] = b
+			} else {
+				if sMap[strList[index]] != b {
+					return false
+				}
+			}
+		} else {
+			if pMap[b] != strList[index] {
+				return false
+			}
+		}
+	}
+	return true
+}
+
+```