1814. 统计一个数组中好对子的数目

Author: w2xi

README.md

@@ -175,6 +175,7 @@
 |1780|[判断一个数字是否可以表示成三的幂的和](https://leetcode.cn/problems/check-if-number-is-a-sum-of-powers-of-three/)|[JavaScript](./algorithms/check-if-number-is-a-sum-of-powers-of-three.js)|Medium|
 |1807|[替换字符串中的括号内容](https://leetcode.cn/problems/evaluate-the-bracket-pairs-of-a-string/)|[JavaScript](./algorithms/evaluate-the-bracket-pairs-of-a-string.js)|Medium|
 |1813|[句子相似性 III](https://leetcode.cn/problems/sentence-similarity-iii/)|[JavaScript](./algorithms/sentence-similarity-iii.js)|Medium|
+|1814|[统计一个数组中好对子的数目](https://leetcode.cn/problems/count-nice-pairs-in-an-array/)|[JavaScript](./algorithms/count-nice-pairs-in-an-array.js)|Medium|
 |1832|[判断句子是否为全字母句](https://leetcode.cn/problems/check-if-the-sentence-is-pangram/)|[JavaScript](./algorithms/check-if-the-sentence-is-pangram.js)|Easy|
 |1945|[字符串转化后的各位数字之和](https://leetcode.cn/problems/sum-of-digits-of-string-after-convert/)|[JavaScript](./algorithms/sum-of-digits-of-string-after-convert.js)|Easy|
 |2032|[至少在两个数组中出现的值](https://leetcode.cn/problems/two-out-of-three/)|[JavaScript](./algorithms/two-out-of-three.js)|Easy|

algorithms/count-nice-pairs-in-an-array.js

@@ -0,0 +1,39 @@
+/**
+ * 1814. 统计一个数组中好对子的数目
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countNicePairs = function (nums) {
+  const rev = (num) => {
+    let result = 0;
+    while (num) {
+      result = result * 10 + (num % 10);
+      num = Math.floor(num / 10);
+    }
+    return result;
+  };
+  
+  // nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])
+  // <=> 等价于
+  // nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])
+
+  // 主要思路 (by 评论区大佬)：
+  // 假设：以 nums[i] - rev(nums[i]) 为整体出现的次数为N，之后假想将这 N 个数放入一个数组中 L
+  // 则： `好的` 就相当于这个数组 L 的所有长度为 2 的子集，而所求个数即为符合条件的子集个数
+  // 易知：长度为 N 的数组 L，其长度为 2 的子集个数为 N * (N - 1) / 2
+
+  let result = 0;
+  const mod = 1e9 + 7;
+  const map = {};
+
+  // 0 + 1 + 2 + ... + (n - 1)
+  // =>
+  // Sum = n * (n - 1) / 2
+
+  for (let i = 0; i < nums.length; i++) {
+    const key = nums[i] - rev(nums[i]);
+    result = (result + (map[key] || 0)) % mod;
+    map[key] = (map[key] || 0) + 1;
+  }
+  return result;
+};