add: 统计出现过一次的公共字符串

README.md

@@ -202,6 +202,10 @@ 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
 
 ### 数组/队列/集合/映射
 
+- [统计出现过一次的公共字符串](src/array/count_common_words_with_one_occurrence.cpp)  [数组, 哈希表, 字符串, 计数]
+
+  - LeetCode 2085. 统计出现过一次的公共字符串 <https://leetcode.cn/problems/count-common-words-with-one-occurrence>
+
 - [回旋镖的数量](src/array/number_of_boomerangs.cpp)  [数组, 哈希表, 数学]
 
   - LeetCode 447. 回旋镖的数量 <https://leetcode.cn/problems/number-of-boomerangs>

src/array/count_common_words_with_one_occurrence.cpp

@@ -0,0 +1,26 @@
+// 统计出现过一次的公共字符串
+// https://leetcode.cn/problems/count-common-words-with-one-occurrence
+// INLINE  ../../images/array/count_common_words_with_one_occurrence.jpeg
+
+#include <headers.hpp>
+
+class Solution {
+public:
+  int countWords(vector<string> &words1, vector<string> &words2) {
+    unordered_map<string, int> map1, map2;
+    for (const string &w : words1) {
+      ++map1[w];
+    }
+    for (const string &w : words2) {
+      ++map2[w];
+    }
+
+    int res = 0;
+    for (const auto &[w, cnt1] : map1) {
+      if (cnt1 == 1 && map2[w] == 1) {
+        ++res;
+      }
+    }
+    return res;
+  }
+};

test/array/count_common_words_with_one_occurrence_test.cpp

@@ -0,0 +1,32 @@
+#include <array/count_common_words_with_one_occurrence.cpp>
+
+TEST(统计出现过一次的公共字符串, countWords) {
+  Solution solution;
+  // 示例 1：
+  // 输入：words1 = ["leetcode","is","amazing","as","is"], words2 =
+  // ["amazing","leetcode","is"] 输出：2 解释：
+  // - "leetcode" 在两个数组中都恰好出现一次，计入答案。
+  // - "amazing" 在两个数组中都恰好出现一次，计入答案。
+  // - "is" 在两个数组中都出现过，但在 words1 中出现了 2 次，不计入答案。
+  // - "as" 在 words1 中出现了一次，但是在 words2 中没有出现过，不计入答案。
+  // 所以，有 2 个字符串在两个数组中都恰好出现了一次。
+  vector<string> words1 = {"leetcode", "is", "amazing", "as", "is"};
+  vector<string> words2 = {"amazing", "leetcode", "is"};
+  EXPECT_EQ(solution.countWords(words1, words2), 2);
+
+  // 示例 2：
+  // 输入：words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]
+  // 输出：0
+  // 解释：没有字符串在两个数组中都恰好出现一次。
+  words1 = {"b", "bb", "bbb"};
+  words2 = {"a", "aa", "aaa"};
+  EXPECT_EQ(solution.countWords(words1, words2), 0);
+
+  // 示例 3：
+  // 输入：words1 = ["a","ab"], words2 = ["a","a","a","ab"]
+  // 输出：1
+  // 解释：唯一在两个数组中都出现一次的字符串是 "ab" 。
+  words1 = {"a", "ab"};
+  words2 = {"a", "a", "a", "ab"};
+  EXPECT_EQ(solution.countWords(words1, words2), 1);
+}

test/lib/lib_test.cpp

@@ -1,4 +1,4 @@
-// 执行编译时间：2024-01-11 11:49:55
+// 执行编译时间：2024-01-12 11:35:49
 #include <gtest/gtest.h>
 #include <lib.hpp>