add: 统计重复个数

README.md

@@ -152,6 +152,10 @@ C++标准库提供的数据结构实在是太多了，参考[C++标准库头文
 
 ### 字符串
 
+- [统计重复个数](src/string/count_the_repetitions.cpp)  [字符串, 动态规划]
+
+  - LeetCode 466. 统计重复个数 <https://leetcode.cn/problems/count-the-repetitions>
+
 - [字典序最小回文串](src/string/lexicographically_smallest_palindrome.cpp)  [贪心, 双指针, 字符串]
 
   - LeetCode 2697. 字典序最小回文串 <https://leetcode.cn/problems/lexicographically-smallest-palindrome>

src/string/count_the_repetitions.cpp

@@ -0,0 +1,48 @@
+// 统计重复个数
+// https://leetcode.cn/problems/count-the-repetitions
+// INLINE  ../../images/string/count_the_repetitions.jpeg
+
+#include <headers.hpp>
+
+class Solution {
+public:
+  int getMaxRepetitions(string s1, int n1, string s2, int n2) {
+    int len1 = s1.length();
+    int len2 = s2.length();
+    int index1 = 0;
+    int index2 = 0;
+
+    if (len1 == 0 || len2 == 0 || len1 * n1 < len2 * n2) {
+      return 0;
+    }
+
+    unordered_map<int, int> map1;
+    unordered_map<int, int> map2;
+    int ans = 0;
+
+    while (index1 / len1 < n1) {
+      if (index1 % len1 == len1 - 1) {
+        if (map1.count(index2 % len2)) {
+          int cycleLen = index1 / len1 - map1[index2 % len2] / len1;
+          int cycleNum = (n1 - 1 - index1 / len1) / cycleLen;
+          int cycleS2Num = index2 / len2 - map2[index2 % len2] / len2;
+
+          index1 += cycleNum * cycleLen * len1;
+          ans += cycleNum * cycleS2Num;
+        } else {
+          map1[index2 % len2] = index1;
+          map2[index2 % len2] = index2;
+        }
+      }
+
+      if (s1[index1 % len1] == s2[index2 % len2]) {
+        if (index2 % len2 == len2 - 1) {
+          ans += 1;
+        }
+        index2 += 1;
+      }
+      index1 += 1;
+    }
+    return ans / n2;
+  }
+};

test/lib/lib_test.cpp

@@ -1,4 +1,4 @@
-// 执行编译时间：2024-01-01 11:39:56
+// 执行编译时间：2024-01-02 09:20:33
 #include <gtest/gtest.h>
 #include <lib.hpp>
 

test/string/count_the_repetitions_test.cpp

@@ -0,0 +1,22 @@
+#include <string/count_the_repetitions.cpp>
+
+TEST(统计重复个数, getMaxRepetitions) {
+  Solution solution;
+  // 示例 1：
+  // 输入：s1 = "acb", n1 = 4, s2 = "ab", n2 = 2
+  // 输出：2
+  string s1 = "acb";
+  int n1 = 4;
+  string s2 = "ab";
+  int n2 = 2;
+  EXPECT_EQ(solution.getMaxRepetitions(s1, n1, s2, n2), 2);
+
+  // 示例 2：
+  // 输入：s1 = "acb", n1 = 1, s2 = "acb", n2 = 1
+  // 输出：1
+  s1 = "acb";
+  n1 = 1;
+  s2 = "acb";
+  n2 = 1;
+  EXPECT_EQ(solution.getMaxRepetitions(s1, n1, s2, n2), 1);
+}