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leetcode update
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@ -13899,3 +13899,207 @@ public:
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};
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```
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## day210 2024-09-24
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### 3043. Find the Length of the Longest Common Prefix
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You are given two arrays with positive integers arr1 and arr2.
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A prefix of a positive integer is an integer formed by one or more of its digits, starting from its leftmost digit. For example, 123 is a prefix of the integer 12345, while 234 is not.
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A common prefix of two integers a and b is an integer c, such that c is a prefix of both a and b. For example, 5655359 and 56554 have a common prefix 565 while 1223 and 43456 do not have a common prefix.
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You need to find the length of the longest common prefix between all pairs of integers (x, y) such that x belongs to arr1 and y belongs to arr2.
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Return the length of the longest common prefix among all pairs. If no common prefix exists among them, return 0.
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### 题解
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题目中的数组要求求最长公共前缀，考虑到每个数组中的不同数字之间也可能存在公共前缀，为了避免对字符串进行重复遍历，可以使用字典树。字典树充分利用了数字每一位的信息，非常适用于类似本题目的求前缀等场景下，还有一道经典题目是求数组中任意两个数字的最大异或和，也可以使用字典树求解，利用异或运算异或两次后会回到数字本身的特性，用数字的二进制形式构造字典树，充分利用每个二进制位的信息，通过贪心尽量将每一位置为1，最终找到最大异或和，可参考[P4551 最长异或路径](https://www.luogu.com.cn/problem/P4551)
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本题在对两个数组构建好两棵字典树后，对字典树同时进行dfs，依次选取每一层中相同的字符对应的子节点进行遍历，直到在某一层无法找到共同字符为止。用一个变量保存遍历到的树的最大深度即为最长公共前缀的长度。
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### 代码
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```cpp
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#pragma GCC optimize("O3", "unroll-loops")
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const int TREESIZE = 10;
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struct TrieNode{
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TrieNode *children[TREESIZE];
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bool isEndOfWord;
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TrieNode(): isEndOfWord(false){
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for (int i = 0;i<TREESIZE;++i){
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children[i] = nullptr;
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}
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}
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};
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class Trie{
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public:
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TrieNode *root;
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Trie(){root = new TrieNode();}
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void insert(const string& word){
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TrieNode* node = root;
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for(char c : word){
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if (node->children[c-'0'] == nullptr){
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node->children[c-'0'] = new TrieNode();
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}
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node = node->children[c-'0'];
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}
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node->isEndOfWord = true;
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}
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~Trie() {
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function<void(TrieNode*)> deleteTrie = [&](TrieNode* node) {
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if(node == nullptr) return;
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for(int i = 0; i < TREESIZE; ++i){
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if(node->children[i] != nullptr){
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deleteTrie(node->children[i]);
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}
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}
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delete node;
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};
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deleteTrie(root);
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};
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};
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class Solution {
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public:
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int longestCommonPrefix(vector<int>& arr1, vector<int>& arr2) {
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ios::sync_with_stdio(false);
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Trie *trie1 = new Trie();
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for (int num : arr1){
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string numstr = to_string(num);
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trie1->insert(numstr);
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}
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Trie *trie2 = new Trie();
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for (int num : arr2){
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string numstr = to_string(num);
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trie2->insert(numstr);
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}
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int result = 0;
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dfs(trie1->root, trie2->root, 0, result);
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return result;
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}
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void dfs(TrieNode* root1,TrieNode* root2, int depth, int& result){
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result = max(result, depth);
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for(int i=0;i<TREESIZE;i++){
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if (root1->children[i] != nullptr && root2->children[i] != nullptr){
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cout<< i;
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dfs(root1->children[i],root2->children[i],depth+1,result);
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}
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}
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}
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};
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```
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## day211 2024-09-25
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### 2416. Sum of Prefix Scores of strings
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You are given an array words of size n consisting of non-empty strings.
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We define the score of a string word as the number of strings words[i] such that word is a prefix of words[i].
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For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since "ab" is a prefix of both "ab" and "abc".
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Return an array answer of size n where answer[i] is the sum of scores of every non-empty prefix of words[i].
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Note that a string is considered as a prefix of itself.
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### 题解
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本题是一道难题，但像这种要多次计算字符串前缀的问题我们已经熟悉了，为了避免重复遍历字符串可以使用字典树trie。在这个问题上思考一下为什么用字典树更好，如果我们用map将某个前缀作为key，其出现的次数作为value。这样对于有包含关系的前缀就会产生大量重复，同时没能充分利用前缀自身自带的字符的先后关系（字符串ab，a是在b前面且与b紧密相邻的，用字典树则将ab相邻和a在b前面这两种字符关系都表示了出来）。
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熟悉字典树后，本题只需对每个字符串构建字典树，对字典树的节点做一些修改，节点中加入表示被访问到次数的count变量，无需记录节点是否为单词的结尾。构建好后对每个字符串，访问字典树，并将路径上所有节点的count相加即得最终的sum。
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### 代码
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```cpp
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const int ALPHABET_SIZE = 26;
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struct TrieNode {
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TrieNode* children[ALPHABET_SIZE];
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int count;
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TrieNode() : count(0) {
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for(int i = 0; i < ALPHABET_SIZE; ++i){
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children[i] = nullptr;
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}
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}
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};
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class Trie {
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public:
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TrieNode* root;
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Trie() { root = new TrieNode(); }
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void insert(const string &word) {
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TrieNode* node = root;
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for(char c : word){
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int index = c - 'a';
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if(index < 0 || index >= ALPHABET_SIZE){
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continue;
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}
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if(node->children[index] == nullptr){
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node->children[index] = new TrieNode();
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}
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node = node->children[index];
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node->count++;
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}
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}
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int count(const string &word){
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int ret = 0;
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TrieNode* node = root;
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for(char c : word){
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int index = c - 'a';
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node = node->children[index];
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ret += node->count;
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}
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return ret;
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}
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~Trie() {
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function<void(TrieNode*)> deleteTrie = [&](TrieNode* node) {
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if(node == nullptr) return;
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for(int i = 0; i < ALPHABET_SIZE; ++i){
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if(node->children[i] != nullptr){
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deleteTrie(node->children[i]);
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}
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}
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delete node;
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};
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deleteTrie(root);
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}
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};
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class Solution {
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public:
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vector<int> sumPrefixScores(vector<string>& words) {
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Trie *trie = new Trie();
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for (string word : words){
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trie->insert(word);
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}
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vector<int> result;
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for (string word : words){
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result.push_back(trie->count(word));
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}
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return result;
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}
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};
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```
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