Merge branch 'main' of github.com:w2xi/leetcode into main

Author: w2xi

README.md

@@ -238,6 +238,7 @@
 |2335|[装满杯子需要的最短总时长](https://leetcode.cn/problems/minimum-amount-of-time-to-fill-cups/)|[JavaScript](./algorithms/minimum-amount-of-time-to-fill-cups.js)|Easy|
 |2341|[数组能形成多少数对](https://leetcode.cn/problems/maximum-number-of-pairs-in-array/)|[JavaScript](./algorithms/maximum-number-of-pairs-in-array.js)|Easy|
 |2351|[第一个出现两次的字母](https://leetcode.cn/problems/first-letter-to-appear-twice/)|[JavaScript](./algorithms/first-letter-to-appear-twice.js)|Easy|
+|2357|[使数组中所有元素都等于零](https://leetcode.cn/problems/make-array-zero-by-subtracting-equal-amounts/)|[JavaScript](./algorithms/make-array-zero-by-subtracting-equal-amounts.js)|Easy|
 |6354|[找出数组的串联值](https://leetcode.cn/problems/find-the-array-concatenation-value/)|[JavaScript](./algorithms/find-the-array-concatenation-value.js)|Easy|
 |6362|[合并两个二维数组 - 求和法](https://leetcode.cn/problems/merge-two-2d-arrays-by-summing-values/)|[JavaScript]()|Easy|
 |面试题 04.12|[面试题 04.12. 求和路径](https://leetcode.cn/problems/paths-with-sum-lcci/)|[JavaScript](./algorithms/paths-with-sum-lcci.js)|Medium|

algorithms/make-array-zero-by-subtracting-equal-amounts.js

@@ -0,0 +1,45 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumOperations = function (nums) {
+  // 排序 + 模拟
+  nums.sort((a, b) => a - b);
+  let count = 0;
+  // 0 1 3 5 5 -> 0 0 2 4 4 -> 0 0 0 2 2 -> 0 0 0 0 0
+  for (let i = 0; i < nums.length; i++) {
+    let curr = nums[i];
+    if (nums[i] !== 0) {
+      for (let j = i; j < nums.length; j++) {
+        nums[j] -= curr;
+      }
+      count++;
+    }
+  }
+  return count;
+};
+
+// 其实就是找不重复数字的个数
+var minimumOperations = function (nums) {
+  nums.sort((a, b) => a - b);
+  let count = 0;
+
+  for (let i = 0; i < nums.length; i++) {
+    if (nums[i] !== 0 && nums[i] !== nums[i + 1]) {
+      count++;
+    }
+  }
+  return count;
+}
+
+var minimumOperations = function (nums) {
+  const set = new Set();
+
+  for (let i = 0; i < nums.length; i++) {
+    if (nums[i] !== 0) {
+      set.add(nums[i]);
+    }
+  }
+
+  return set.size;
+}