138

Author: tech-cow

README.md

@@ -37,3 +37,7 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |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 | ||
 |19|[Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/#/description)| [Python](./linkedlist/removeNthFromEnd.py) | _O(n)_| _O(1)_  | Medium | ||
+
+
+## 4/16 Tasks (LinkedList Medium)
+|138|[Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer/#/description)| [Python](./linkedlist/copyRandomList.py) | _O(n)_| _O(n)_  | Medium | ||

linkedlist/copyRandomList.py

@@ -0,0 +1,43 @@
+# 138. Copy List with Random Pointer
+# Author: Yu Zhou
+
+# A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
+# Return a deep copy of the list
+# Definition for singly-linked list with a random pointer.
+# class RandomListNode(object):
+#     def __init__(self, x):
+#         self.label = x
+#         self.next = None
+#         self.random = None
+
+# 思路：
+# 这题其实就是拷贝所有的链表到一个新的
+# 唯一的难点是如何在拷贝玩所有链表之后，还能够将Random的指针指对
+# 所以这里用的方法是用Hash，然后第一次遍历用来建新的Node
+# 第二次遍历用来对Hash里面新建的Node进行指针的赋值
+
+class Solution(object):
+    def copyRandomList(self, head):
+        """
+        :type head: RandomListNode
+        :rtype: RandomListNode
+        """
+        dic = dict()
+        m = n = head
+
+        #第一次循环用来创建所有原本链表里的Node，并且对他们进行赋值
+        #需要搞懂的是，这一次循环，新的链表是一个一个单独的Node，他们的指针都还没有赋值
+        while m:
+            dic[m] = RandomListNode(m.label)
+            m = m.next
+
+        #第二次循环，用来对这一个一个独立的Node进行他们的指针赋值，其中包括每个链接下一个Node的指针
+        #以及这道题的另一个Random的指针
+        while n:
+            #这里dic[n].next，dic[n]就是key "n" 相对应的val，也就是之前创建的，RandomListNode(m.label)
+            #然后我们对这个新创建的独立Node，设置他的next pointer
+            # dic.get(n.next）里面的n.next是reference原List里面的Node的next，不要搞混了
+            dic[n].next = dic.get(n.next)
+            dic[n].random = dic.get(n.random)
+            n = n.next
+        return dic.get(head)