0061. Rotate List

Author: coderbean

markdown/0061. Rotate List.md

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+### [61\. Rotate List](https://leetcode.com/problems/rotate-list/)
+
+Difficulty: **Medium**
+
+
+Given a linked list, rotate the list to the right by _k_ places, where _k_ is non-negative.
+
+**Example 1:**
+
+```
+Input: 1->2->3->4->5->NULL, k = 2
+Output: 4->5->1->2->3->NULL
+Explanation:
+rotate 1 steps to the right: 5->1->2->3->4->NULL
+rotate 2 steps to the right: 4->5->1->2->3->NULL
+```
+
+**Example 2:**
+
+```
+Input: 0->1->2->NULL, k = 4
+Output: 2->0->1->NULL
+Explanation:
+rotate 1 steps to the right: 2->0->1->NULL
+rotate 2 steps to the right: 1->2->0->NULL
+rotate 3 steps to the right: 0->1->2->NULL
+rotate 4 steps to the right: 2->0->1->NULL```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ * int val;
+ * ListNode next;
+ * ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode rotateRight(ListNode head, int k) {
+        if (head == null || head.next == null) {
+            return head;
+        }
+        int length = 0;
+        ListNode index = head;
+        while (index.next != null) {
+            length++;
+            index = index.next;
+        }
+        length++;
+        index.next = head; // 构建成循环链表，index 指向 head 的前一个节点
+        k = k % length;
+        for (int i = 0; i < length - k; i++) {
+            index = head;
+            head = head.next;
+        }
+        index.next = null; // 破除循环链表，添加结束 null 值
+        return head;
+    }
+}
+```
+![](http://ww1.sinaimg.cn/large/006tNc79ly1g50orael7nj31b80qsaeq.jpg)

src/main/java/leetcode/_61_/Main.java

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+package leetcode._61_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println( solution.isValid(""));
+    }
+}
+

src/main/java/leetcode/_61_/Solution.java

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+package leetcode._61_;
+
+import leetcode.common.ListNode;
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ * int val;
+ * ListNode next;
+ * ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode rotateRight(ListNode head, int k) {
+        if (head == null || head.next == null) {
+            return head;
+        }
+        int length = 0;
+        ListNode index = head;
+        while (index.next != null) {
+            length++;
+            index = index.next;
+        }
+        length++;
+        index.next = head; // 构建成循环链表，index 指向 head 的前一个节点
+        k = k % length;
+        for (int i = 0; i < length - k; i++) {
+            index = head;
+            head = head.next;
+        }
+        index.next = null; // 破除循环链表，添加结束 null 值
+        return head;
+    }
+}

src/main/java/leetcode/_61_/solution.md

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+### [61\. Rotate List](https://leetcode.com/problems/rotate-list/)
+
+Difficulty: **Medium**
+
+
+Given a linked list, rotate the list to the right by _k_ places, where _k_ is non-negative.
+
+**Example 1:**
+
+```
+Input: 1->2->3->4->5->NULL, k = 2
+Output: 4->5->1->2->3->NULL
+Explanation:
+rotate 1 steps to the right: 5->1->2->3->4->NULL
+rotate 2 steps to the right: 4->5->1->2->3->NULL
+```
+
+**Example 2:**
+
+```
+Input: 0->1->2->NULL, k = 4
+Output: 2->0->1->NULL
+Explanation:
+rotate 1 steps to the right: 2->0->1->NULL
+rotate 2 steps to the right: 1->2->0->NULL
+rotate 3 steps to the right: 0->1->2->NULL
+rotate 4 steps to the right: 2->0->1->NULL```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ * int val;
+ * ListNode next;
+ * ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode rotateRight(ListNode head, int k) {
+        if (head == null || head.next == null) {
+            return head;
+        }
+        int length = 0;
+        ListNode index = head;
+        while (index.next != null) {
+            length++;
+            index = index.next;
+        }
+        length++;
+        index.next = head; // 构建成循环链表，index 指向 head 的前一个节点
+        k = k % length;
+        for (int i = 0; i < length - k; i++) {
+            index = head;
+            head = head.next;
+        }
+        index.next = null; // 破除循环链表，添加结束 null 值
+        return head;
+    }
+}
+```
+![](http://ww1.sinaimg.cn/large/006tNc79ly1g50orael7nj31b80qsaeq.jpg)