rotate list medium

Array/RotateList.py

@@ -0,0 +1,98 @@
+"""
+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
+
+
+旋转链表。 k 非负。
+
+k 超过链表的最大长度也可。
+
+思路：
+
+过一遍链表长度，k % length 取个模，防止 k 超级大。
+
+之后 slow fast 两个，fast 先走 k 个，然后 slow 与 fast 同时走，走到最后 slow.next 即为从后到前 k 个的起点。
+
+剩下的就是把原来的尾置换到前。
+
+下面这个可以优化下，不过就测试来说已经可以了。
+
+beat 100%
+24ms ~ 36ms
+
+测试地址：
+https://leetcode.com/problems/rotate-list/description/
+
+"""
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+
+    def rotateRight(self, head, k):
+        """
+        :type head: ListNode
+        :type k: int
+        :rtype: ListNode
+        """
+        if not k or not head:
+            return head
+
+        def getLength(node):
+            length = 0
+
+            while node:
+                node = node.next
+                length += 1
+
+            return length
+
+        length = getLength(head)
+        k = k % length
+
+        slow = head
+        fast = head
+
+        while k > 0:
+            fast = fast.next
+
+            k -= 1
+
+        while fast.next:
+            slow = slow.next
+            fast = fast.next
+
+        rotate_head = slow.next
+
+        if not rotate_head:
+            return head
+
+        slow.next = None
+
+        _rotate_head = rotate_head
+        while _rotate_head.next:
+            _rotate_head = _rotate_head.next
+
+        _rotate_head.next = head
+
+        return rotate_head
+