Merge branch 'main' of https://github.com/tkzzzzzz6/Algorithm_beginner_learning_notes

Author: tkzzzzzz6

leetcode/problems/3043.find-the-length-of-the-longest-common-prefix.cpp

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+/*
+ * @lc app=leetcode.cn id=3043 lang=cpp
+ * @lcpr version=30204
+ *
+ * [3043] 最长公共前缀的长度
+ */
+
+
+// @lcpr-template-start
+using namespace std;
+#include <algorithm>
+#include <array>
+#include <bitset>
+#include <climits>
+#include <deque>
+#include <functional>
+#include <iostream>
+#include <list>
+#include <queue>
+#include <stack>
+#include <tuple>
+#include <unordered_map>
+#include <unordered_set>
+#include <utility>
+#include <vector>
+// @lcpr-template-end
+// @lc code=start
+class Solution {
+public:
+    int longestCommonPrefix(vector<int>& arr1, vector<int>& arr2) {
+        unordered_set<string> st;
+        for(auto x:arr1){
+            string s = to_string(x);
+            for(int i = 1;i<=s.size();++i){ //枚举 s 的前缀,并压入哈希
+                st.insert(s.substr(0,i));
+            }
+        }
+
+        int ans = 0;
+        for(auto x:arr2){
+            string s = to_string(x);
+            for(int i = 1;i<=s.size();++i){ //寻找 arr2 与 arr1 中共同有的前缀,并统计最大值
+                if(!st.contains(s.substr(0,i)))break; 
+                ans = max(ans,i);
+            }
+        }
+        return ans;
+    }
+};
+// @lc code=end
+
+
+
+/*
+// @lcpr case=start
+// [1,10,100]\n[1000]\n
+// @lcpr case=end
+
+// @lcpr case=start
+// [1,2,3]\n[4,4,4]\n
+// @lcpr case=end
+
+ */

leetcode/problems/3043.find-the-length-of-the-longest-common-prefix.md

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+<!--
+ * @Author: tkzzzzzz6
+ * @Date: 2026-05-21 11:22:00
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-05-21 11:26:23
+-->
+# [3043] 最长公共前缀的长度
+
+## 题目描述
+
+## 思路:哈希
+
+- 时间复杂度:
+- 空间复杂度:
+
+```cpp
+class Solution {
+public:
+    int longestCommonPrefix(vector<int>& arr1, vector<int>& arr2) {
+        unordered_set<string> st;
+        for(auto x:arr1){
+            string s = to_string(x);
+            for(int i = 1;i<=s.size();++i){ //枚举 s 的前缀,并压入哈希
+                st.insert(s.substr(0,i));
+            }
+        }
+
+        int ans = 0;
+        for(auto x:arr2){
+            string s = to_string(x);
+            for(int i = 1;i<=s.size();++i){ //寻找 arr2 与 arr1 中共同有的前缀,并统计最大值
+                if(!st.contains(s.substr(0,i)))break; 
+                ans = max(ans,i);
+            }
+        }
+        return ans;
+    }
+};
+```