Friday, October 4, 2013

138. Copy List with Random Pointer ----M

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.
The Linked List is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:
  • val: an integer representing Node.val
  • random_index: the index of the node (range from 0 to n-1) where random pointer points to, or null if it does not point to any node.

Example 1:

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
Example 2:

Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]
Example 3:
Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]
Example 4:
Input: head = []
Output: []
Explanation: Given linked list is empty (null pointer), so return null.

Constraints:
  • -10000 <= Node.val <= 10000
  • Node.random is null or pointing to a node in the linked list.
  • Number of Nodes will not exceed 1000.

A:

啊啊啊啊啊啊,终于明白了。这个题目考点在于 random有可能是指向后面的节点。
如果仅仅是通常的copy的话,无法当即设置random 指针。 --------因此可以用map
/*
// Definition for a Node.
class Node {
public:
int val;
Node* next;
Node* random;

Node(int _val) {
val = _val;
next = NULL;
random = NULL;
}
};
*/

class Solution {
public:
Node* copyRandomList(Node* head) {
Node header(0);
unordered_map<Node*, Node*> map;
auto runner = head;
auto tail = &header;
while (runner) {
tail->next = new Node(runner->val);
tail = tail->next;
map[runner] = tail; // DO NOT use insert
runner = runner->next;
}
// now another round, setup the random pointer
runner = head;
auto newRunner = header.next;
while (runner) {
newRunner->random = map[runner->random];
runner = runner->next;
newRunner = newRunner->next;
}
return header.next;
}
};


Learned:

-------------------------------Third time --------------------
travel the list two pass ,
Pass 1:  create the list and insert right after that node,
Pass 2:  set up the random node
Pass 3:  detach the copied node list.

(As Pass 2 and Pass 3 can do together, we thus need only two pass.)
Correction !!!!!!!!!!!
Pass 2 and Pass 3 can NOT do together.






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