leetcode update

content/posts/leetcode.md

@@ -7003,3 +7003,86 @@ func isNStraightHand(hand []int, groupSize int) bool {
 
 }
 ```
+
+## day102 2024-06-07
+
+### 648. Replace Words
+
+In English, we have a concept called root, which can be followed by some other word to form another longer word - let's call this word derivative. For example, when the root "help" is followed by the word "ful", we can form a derivative "helpful".
+
+Given a dictionary consisting of many roots and a sentence consisting of words separated by spaces, replace all the derivatives in the sentence with the root forming it. If a derivative can be replaced by more than one root, replace it with the root that has the shortest length.
+
+Return the sentence after the replacement.
+
+![0607mDCIwsOF5xfV](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0607mDCIwsOF5xfV.png)
+
+### 题解
+
+读懂题是第一步, 本题要将句子中所有的单词替换为在词典中对应的它的前缀形式(如果字典中存在这个单词的前缀的话), 最直接的思路就是遍历句子中的所有单词, 将每个单词依次和词典中的所有单词比较, 看词典中是否存在它的前缀, 但这种方法无疑效率极低, 这个场景和我们日常查词典有些相似, 想想我们日常在词典中查一个单词是如何快速定位的, 当然是按照字母顺序在词典中先定位第一个字母的范围, 再定位第二个字母的范围...,最后找到需要的单词. 现在只要把这种方法表示出来就可以大大加快寻找前缀的速度. 这样就可以构造一棵词典树, 树的边表示不同的字母, 节点可以用来标定是否是一个可行单词. 再去查找某个单词的前缀的时候, 只需要去树中执行bfs找到最短的可行前缀即可.
+
+### 代码
+
+```go
+type TreeNodes struct{
+    node [26]*TreeNodes
+    over bool
+}
+
+func construct(dic []string) *TreeNodes{
+    root := TreeNodes{}
+    var point *TreeNodes
+    for _, str := range dic{
+        point = &root
+        for i,_ := range str{
+            if point.node[str[i]-'a']==nil{
+                newnode := &TreeNodes{}
+                point.node[str[i]-'a'] = newnode
+                point = newnode
+            }else{
+                point = point.node[str[i]-'a']
+            }
+        }
+        point.over = true
+    }
+    return &root
+}
+
+func findWord(root *TreeNodes, word string)(bool, string){
+    var point *TreeNodes
+    point = root
+    prefix := []byte{}
+    for i,_ := range word{
+        if point.node[word[i]-'a'] == nil{
+            return false,""
+        }else{
+            prefix = append(prefix, word[i])
+            point = point.node[word[i]-'a']
+            if point.over{
+                return true, string(prefix)
+            }
+        }
+    }
+    return false,""
+}
+
+func replaceWords(dictionary []string, sentence string) string {
+    root := construct(dictionary)
+    sentenceArray := strings.Split(sentence, " ")
+    exist, str := false, ""
+    for i, word := range sentenceArray{
+        exist, str = findWord(root, word)
+        if exist{
+            sentenceArray[i] = str
+        }else{
+            sentenceArray[i] = word
+        }
+    }
+    return strings.Join(sentenceArray, " ")
+}
+
+
+```
+
+### 总结
+
+注意直接使用"+"在go中进行字符串连接是非常耗时的, 因此可以将原句子分割后得到字符串数组, 查询是否存在单词的可行前缀并直接替换对应数组中的字符串, 最后再调用strings.Join将数组中的所有字符串使用空格连接, 可以自行测试如果对数组中每个字符串在查找其可行前缀后都使用"+"连接到结果字符串上耗时远远大于这种方案.