提交24和25题

yllg · yllg · commit b2d91d65dc92 · 2018-03-23T19:08:45.000+08:00
diff --git a/024.交换相邻节点/024 solution.html b/024.交换相邻节点/024 solution.html
@@ -0,0 +1,74 @@
+<!DOCTYPE html>
+<html lang="en">
+
+<head>
+    <meta charset="UTF-8">
+    <meta name="viewport" content="width=device-width, initial-scale=1.0">
+    <meta http-equiv="X-UA-Compatible" content="ie=edge">
+    <title>Document</title>
+</head>
+
+<body>
+    <script>
+        // Source : https://leetcode.com/problems/swap-nodes-in-pairs/
+        // Author : 悬笔e绝
+        // Date   : 2018-03-23
+       
+        /**
+         * @param {ListNode} head
+         * @return {ListNode}
+         */
+        var swapPairs = function (head) {
+            if (!head)
+                return null;
+            var arr = [];
+            //把链表的每个节点放到arr中，且next都置null
+            while (head) {
+                var next = head.next;
+                head.next = null;
+                arr.push(head);
+                head = next;
+            }
+
+            var len = arr.length;
+            //数组每两个一组交换顺序
+            for (var i = 0; i < len; i += 2) {
+                var a = arr[i];
+                var b = arr[i + 1];
+                //数组长度为奇数的情况
+                if (!b)
+                    continue;
+                //互换
+                arr[i] = b;
+                arr[i + 1] = a;
+            }
+            //把两两互换之后的节点串起来~
+            for (var i = 0; i < len - 1; i++)
+                arr[i].next = arr[i + 1];
+            //返回头节点
+            return arr[0];
+        };
+
+        // Definition for singly-linked list.
+        function ListNode(val) {
+            this.val = val;
+            this.next = null;
+        }
+
+        //测试 1->2->3->4->5
+        var N5 = new ListNode(5);
+        var N4 = new ListNode(4);
+        N4.next = N5;
+        var N3 = new ListNode(3);
+        N3.next = N4;
+        var N2 = new ListNode(2);
+        N2.next = N3;
+        var N1 = new ListNode(1);
+        N1.next = N2;
+
+        console.log(swapPairs(N1));
+
+    </script>
+</body>
+
+</html>
diff --git a/025.每K个1组翻转链表/025 solution.html b/025.每K个1组翻转链表/025 solution.html
@@ -0,0 +1,80 @@
+<!DOCTYPE html>
+<html lang="en">
+
+<head>
+    <meta charset="UTF-8">
+    <meta name="viewport" content="width=device-width, initial-scale=1.0">
+    <meta http-equiv="X-UA-Compatible" content="ie=edge">
+    <title>Document</title>
+</head>
+
+<body>
+    <script>
+        // Source : https://leetcode.com/problems/reverse-nodes-in-k-group/
+        // Author : 悬笔e绝
+        // Date   : 2018-03-23
+
+        /**
+         * @param {ListNode} head
+         * @param {number} k
+         * @return {ListNode}
+         */
+
+         //其实就是24的进阶版，关键使用了数组的reserve倒置方法
+        var reverseKGroup = function (head, k) {
+            var ans = [];
+
+            while (head) {
+                ans.push(new ListNode(head.val));
+                head = head.next;
+            }
+
+            if (!ans.length)
+                return null;
+
+            var len = ans.length;
+            var res = [];
+
+            for (var i = 0; i < len; i += k) {
+                var tmp;
+                //如果超出长度，截取最后一段，且不用倒置
+                if (i + k > len)
+                    tmp = ans.slice(i, len);
+                else {
+                    //没有超出长度就截取并倒置，比如0到3，截取前3个
+                    tmp = ans.slice(i, i + k);
+                    tmp.reverse();
+                }
+                //把tmp不断的加到res中
+                Array.prototype.push.apply(res, tmp);
+            }
+            //把res中的节点串起来
+            for (var i = 0, len = res.length - 1; i < len; i++)
+                res[i].next = res[i + 1];
+
+            return res[0];
+        };
+
+        // Definition for singly-linked list.
+        function ListNode(val) {
+            this.val = val;
+            this.next = null;
+        }
+
+        //测试 1->2->3->4->5
+        var N5 = new ListNode(5);
+        var N4 = new ListNode(4);
+        N4.next = N5;
+        var N3 = new ListNode(3);
+        N3.next = N4;
+        var N2 = new ListNode(2);
+        N2.next = N3;
+        var N1 = new ListNode(1);
+        N1.next = N2;
+
+        console.log(reverseKGroup(N1,3));
+
+    </script>
+</body>
+
+</html>
