feat: add solutions to lc problem: No.0324

No.0324.Wiggle Sort II

Author: yanglbme

solution/0300-0399/0324.Wiggle Sort II/README.md

@@ -233,6 +233,37 @@ func wiggleSort(nums []int) {
 }
 ```
 
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {void} Do not return anything, modify nums in-place instead.
+ */
+var wiggleSort = function (nums) {
+    let bucket = new Array(5001).fill(0);
+    for (const v of nums) {
+        bucket[v]++;
+    }
+    const n = nums.length;
+    let j = 5000;
+    for (let i = 1; i < n; i += 2) {
+        while (bucket[j] == 0) {
+            --j;
+        }
+        nums[i] = j;
+        --bucket[j];
+    }
+    for (let i = 0; i < n; i += 2) {
+        while (bucket[j] == 0) {
+            --j;
+        }
+        nums[i] = j;
+        --bucket[j];
+    }
+};
+```
+
 ### **...**
 
 ```

solution/0300-0399/0324.Wiggle Sort II/README_EN.md

@@ -218,6 +218,37 @@ func wiggleSort(nums []int) {
 }
 ```
 
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {void} Do not return anything, modify nums in-place instead.
+ */
+var wiggleSort = function (nums) {
+    let bucket = new Array(5001).fill(0);
+    for (const v of nums) {
+        bucket[v]++;
+    }
+    const n = nums.length;
+    let j = 5000;
+    for (let i = 1; i < n; i += 2) {
+        while (bucket[j] == 0) {
+            --j;
+        }
+        nums[i] = j;
+        --bucket[j];
+    }
+    for (let i = 0; i < n; i += 2) {
+        while (bucket[j] == 0) {
+            --j;
+        }
+        nums[i] = j;
+        --bucket[j];
+    }
+};
+```
+
 ### **...**
 
 ```

solution/0300-0399/0324.Wiggle Sort II/Solution.js

@@ -0,0 +1,26 @@
+/**
+ * @param {number[]} nums
+ * @return {void} Do not return anything, modify nums in-place instead.
+ */
+var wiggleSort = function (nums) {
+    let bucket = new Array(5001).fill(0);
+    for (const v of nums) {
+        bucket[v]++;
+    }
+    const n = nums.length;
+    let j = 5000;
+    for (let i = 1; i < n; i += 2) {
+        while (bucket[j] == 0) {
+            --j;
+        }
+        nums[i] = j;
+        --bucket[j];
+    }
+    for (let i = 0; i < n; i += 2) {
+        while (bucket[j] == 0) {
+            --j;
+        }
+        nums[i] = j;
+        --bucket[j];
+    }
+};

solution/0900-0999/0952.Largest Component Size by Common Factor/README.md

@@ -63,7 +63,7 @@
 
 **方法一：数学 + 并查集**
 
-利用“试除法”，对 $nums$ 中的每个数 $v$ 分解因数，然后将每个因数 $i$ 与 v 合并，$v / i$ 与 $v$ 合并。此过程用并查集来维护连通分量。
+利用“试除法”，对 $nums$ 中的每个数 $v$ 分解因数，然后将每个因数 $i$ 与 $v$ 合并，$v / i$ 与 $v$ 合并。此过程用并查集来维护连通分量。
 
 最后，遍历 $nums$ 中每个数 $v$，找出所在的连通分量，出现次数最多的连通分量就是所求的答案。