Rotate List

Author: tech-cow

README.md

@@ -35,3 +35,4 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |-----|-------| -------- | ---- | ------|------------|---|-----|
 |83|[Remove Duplicates from Sorted List](https://leetcode.com/problems/remove-duplicates-from-sorted-list/#/description)| [Python](./linkedlist/deleteDuplicates.py) | _O(n)_| _O(1)_  | Easy | ||
 |148|[Sort List](https://leetcode.com/problems/sort-list/#/description)| [Python](./linkedlist/sortList.py) | _O(nlogn)_| _O(1)_  | Medium | ||
+|61|[Rotate List](https://leetcode.com/problems/rotate-list/#/description)| [Python](./linkedlist/rotateRight.py) | _O(n)_| _O(1)_  | Medium | ||

linkedlist/rotateRight.py

@@ -0,0 +1,78 @@
+# Author: Yu Zhou
+# 61. Rotate List
+
+# Given a list, rotate the list to the right by k places, where k is non-negative.
+#
+# For example:
+# Given 1->2->3->4->5->NULL and k = 2,
+# return 4->5->1->2->3->NULL.
+
+# Time: O(n)
+# Space: O(1)
+
+# 思路
+# 先Pad距离，然后用Mod测量周期，最后将之前切断的两节链表链接起来
+
+# ****************
+# Final Solution *
+# ****************
+class Solution(object):
+    def rotateRight(self, head, k):
+        #Edge
+        if not head or not head.next:
+            return head
+
+        cur, count = head, 0
+
+        #Count Length for modular operation
+        while cur:
+            cur = cur.next
+            count += 1
+        mod = k % count
+
+        #Edge2
+        if k == 0 or mod == 0:
+            return head
+
+        slow = fast = head
+        for _ in xrange(mod):
+            fast = fast.next
+
+        # Use fast.next here so fast point doesn't reach Null
+        while fast.next:
+            slow = slow.next
+            fast = fast.next
+
+        newHead = slow.next
+        fast.next = head
+        slow.next = None
+
+        return newHead
+
+# ***************************************
+# The following code is an fail attempt *
+# ***************************************
+class Solution(object):
+    def rotateRight(self, head, k):
+        if not head:
+            return None
+
+        slow = fast = head
+
+        for _ in xrange(k):
+            fast = fast.next
+
+        prev = None
+        while fast:
+            prev = slow
+            slow = slow.next
+            fast = fast.next
+
+        prev.next = None
+        newHead = cur = slow
+
+        while cur.next:
+            cur = cur.next
+
+        cur.next = head
+        return newHead