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docs/Leetcode_Solutions/C++/0061._Rotate_List.md

+# 61. Rotate List
+
+
+
+**<font color=red>难度:Medium<font>**
+
+
+
+## 刷题内容
+
+> 原题连接
+
+
+* https://leetcode.com/problems/rotate-list/
+
+
+
+> 内容描述
+
+
+
+```
+
+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
+
+```
+
+
+
+> 思路1
+
+******- 时间复杂度: O(n)******- 空间复杂度: O(n)******
+
+
+
+这里的n是指链表的长度，我们用k对n的长度取余，因为k大于链表的长度之后就是一个循环。不需要重复操作。
+
+
+
+```cpp
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* rotateRight(ListNode* head, int k) {
+        int len = 0;
+        ListNode* h = head;
+        ListNode* pre;
+        while(h)
+        {
+            ++len;
+            pre = h;
+            h = h ->next;
+        }
+        if(!len)
+            return head;    
+        int count1 = len - k % len;
+        if(count1 == len)
+            return head;
+        h = head;
+        ListNode* tail;
+        while(count1)
+        {
+            tail = h;
+            h = h ->next;
+            --count1;
+        }
+        pre ->next = head;
+        tail ->next = nullptr;
+        return h;
+    }
+};
+```
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