22w6: add 283 js solution

Author: swiftwind0405

README.md

@@ -193,7 +193,7 @@
 |287|[Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number/)  | |Hard|
 |285|[Inorder Successor in BST](https://leetcode.com/problems/inorder-successor-in-bst/) ♥ | |Medium|
 |284|[Peeking Iterator](https://leetcode.com/problems/peeking-iterator/)  | |Medium|
-|283|[Move Zeroes](https://leetcode.com/problems/move-zeroes/)  | |Easy|
+|283|[Move Zeroes](https://leetcode.com/problems/move-zeroes/)  |  [js](./algorithms/moveZeroes/solution.js) |Easy|
 |282|[Expression Add Operators](https://leetcode.com/problems/expression-add-operators/)  | |Hard|
 |279|[Perfect Squares](https://leetcode.com/problems/perfect-squares/) | |Medium|
 |278|[First Bad Version](https://leetcode.com/problems/first-bad-version/)| |Easy|

algorithms/moveZeroe/solution.js

@@ -0,0 +1,46 @@
+// 解法一：双循环，这是很笨的一种实现方式，时间复杂度也很高
+var moveZeroes = function(nums) {
+    let x = 0;
+    while(x < nums.length) {
+        if(nums[x] === 0) {
+            let y = x + 1;
+            while (x < nums.length - 1 && y < nums.length - 1 && nums[y] === 0) {
+                y++;
+            }
+            if (nums[y]) {
+                [nums[x], nums[y]] = [nums[y], nums[x]]
+            }
+        }
+        x++;
+    }
+    return nums;
+};
+
+// 解法二：双循环的第二种解法
+var moveZeroes = function(nums) {
+    let y = 0;
+    // 第一次遍历的时候，y指针记录非0的个数，只要是非0的统统都赋给nums[y]
+    for (let x = 0; x < nums.length; x++) {
+        if(nums[x] !== 0) {
+            nums[y++] = nums[x];
+        }
+    }
+    // 非0元素统计完了，剩下的都是0了
+	// 所以第二次遍历把末尾的元素都赋为0即可
+    for (let x = y; x < nums.length; x++) {
+        nums[x] = 0;
+    }
+    return nums;
+};
+
+// 解法三：快排的思想（利用0，不等于0的都放到0的左边），达到一次循环
+var moveZeroes = function(nums) {
+    let y = 0;
+    for(let x = 0; x < nums.length; x++) {
+        //当前元素!=0，就把其交换到左边，等于0的交换到右边
+        if(nums[x] !== 0) {
+            [nums[x], nums[y++]] = [nums[y], nums[x]];
+        }
+    }
+    return nums;
+};