leetcode update

content/posts/leetcode.md

@@ -14887,3 +14887,113 @@ public:
 };
 
 ```
+
+## day222 2024-10-06
+
+### 1813. Sentence Similarity III
+
+You are given two strings sentence1 and sentence2, each representing a sentence composed of words. A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each word consists of only uppercase and lowercase English characters.
+
+Two sentences s1 and s2 are considered similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. Note that the inserted sentence must be separated from existing words by spaces.
+
+For example,
+
+s1 = "Hello Jane" and s2 = "Hello my name is Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in s1.
+s1 = "Frog cool" and s2 = "Frogs are cool" are not similar, since although there is a sentence "s are" inserted into s1, it is not separated from "Frog" by a space.
+Given two sentences sentence1 and sentence2, return true if sentence1 and sentence2 are similar. Otherwise, return false.
+
+![1006vIkjQL7Rbg8x](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/1006vIkjQL7Rbg8x.png)
+
+### 题解
+
+本题只能插入一个句子，则要在插入一个句子后将短句变为长句，短句拆分后的前后两部分应分别位于长句的首尾部分（可以拆为空句，这样就是短句完全是长句的开头部分或者结尾部分），则遍历两个句子，匹配开头部分，不匹配时继续向后遍历长句直到可以继续匹配，此时匹配的是短句的后半部分，注意可能存在短句的后半部分的的某部分被重复匹配的情况，即假设短句为aaab，长句可能为aaaaab，因此需要一个指针记录匹配短句后半部分时的起始位置，另外一个指针则进行正常的长短句匹配。
+
+### 代码
+
+```cpp
+class Solution {
+public:
+    bool areSentencesSimilar(string sentence1, string sentence2) {
+        vector<string> s1;
+        vector<string> s2;
+        istringstream iss1(sentence1);
+        istringstream iss2(sentence2);
+        string word;
+        while(iss1 >> word){
+            s1.push_back(word);
+        }
+        while(iss2 >> word){
+            s2.push_back(word);
+        }
+        if (s1.size() > s2.size()){
+            swap(s1,s2);
+        }
+        int start = 0;
+        for (;start<s1.size();start++){
+            if(s1[start] != s2[start]){
+                break;
+            }
+        }
+        int tail = start;
+        if (tail == s1.size()){
+            return true;
+        }
+        int s2ptr = start;
+        // cout << tail << " "<<s2ptr<<" "<<s1.size() << s2.size();
+        for(;s2ptr<s2.size();s2ptr++){
+            if (tail >= s1.size()){
+                tail = start;
+            }
+            if(s2[s2ptr] != s1[tail]){
+                tail = start;
+                if(s2[s2ptr] == s1[tail]){
+                    tail++;
+                }
+            }else{
+                tail++;
+            }
+        }
+        if (tail == s1.size()){
+            return true;
+        }
+        return false;
+    }
+};
+```
+
+### 总结
+
+使用KMP算法应该还可以对字符串匹配的过程做进一步的优化，把每个单词视为kmp中的字母，即可使用同样的方案加速匹配，在这里就不继续改进了。
+
+本题还可采用将短句视为一个双端队列，遍历短句时同时将短句头部和长句头部，短句尾部和长句尾部进行匹配，任意一端匹配成功则弹出一个单词，两端都不匹配的时候结束遍历，判断短句是否已经全部弹出即可得到结果，空则说明可以匹配上，非空则不能匹配上。
+
+```cpp
+class Solution {
+public:
+    bool areSentencesSimilar(string sentence1, string sentence2) {
+        deque<string> v1, v2;
+        stringstream ss(sentence1);
+        string s;
+        while(getline(ss,s,' ')){
+            v1.push_back(s);
+        }
+        stringstream ss2(sentence2);
+        while(getline(ss2,s,' ')){
+            v2.push_back(s);
+        }
+        if(v1.size() > v2.size()) swap(v1,v2);
+        while(v1.size()){
+            if(v1.front() == v2.front()){
+                v1.pop_front();
+                v2.pop_front();
+            } else if(v1.back() == v2.back()){
+                v1.pop_back();
+                v2.pop_back();
+            } else {
+                break;
+            }
+        }
+        return v1.empty();
+    }
+};
+```