138 Copy List with Random Pointer

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.

copy 的题多用hashmap, 难点在于如何让遍历, 如何构建新的节点间的关系.

An intuitive solution is to keep a hash table for each node in the list, via which we just need to iterate the list in 2 rounds respectively to create nodes and assign the values for their random pointers. As a result, the space complexity of this solution is O(N), although with a linear time complexity.

As an optimised solution, we could reduce the space complexity into constant. The idea is to associate the original node with its copy node in a single linked list. In this way, we don't need extra space to keep track of the new nodes.

The algorithm is composed of the follow three steps which are also 3 iteration rounds.

1. Iterate the original list and duplicate each node. The duplicate
   of each node follows its original immediately.
2. Iterate the new list and assign the random pointer for each
   duplicated node.
3. Restore the original list and extract the duplicated nodes.

 

/**
 * Definition for singly-linked list with a random pointer.
 * class RandomListNode {
 *     int label;
 *     RandomListNode next, random;
 *     RandomListNode(int x) { this.label = x; }
 * };
 */
public class Solution {
   public RandomListNode copyRandomList(RandomListNode head) {
        // write your code here
        HashMap<RandomListNode, RandomListNode> map = new HashMap<RandomListNode, RandomListNode>();
        RandomListNode cur = head;
        while (cur != null) {
            RandomListNode copy = new RandomListNode(cur.label);
            map.put(cur, copy);
            cur = cur.next;
        }
        cur = head;
        while (cur != null) {
            map.get(cur).next = map.get(cur.next);
            map.get(cur).random = map.get(cur.random);
            cur = cur.next;
        }
        return map.get(head);
   }

}

通过链表之间的指针指代关系, 要画图, 判空, 在遇到next.next 的情况, 可以用dummyNode, cur = head 返回重新下一个.

The algorithm is composed of the follow three steps which are also 3 iteration rounds.

1. Iterate the original list and duplicate each node. The duplicate
   of each node follows its original immediately.
2. Iterate the new list and assign the random pointer for each
   duplicated node.
3. Restore the original list and extract the duplicated nodes.

public RandomListNode copyRandomList(RandomListNode head) {
	RandomListNode iter = head, next;

	// First round: make copy of each node,
	// and link them together side-by-side in a single list.
	while (iter != null) {
		next = iter.next;

		RandomListNode copy = new RandomListNode(iter.label);
		iter.next = copy;
		copy.next = next;

		iter = next;
	}

	// Second round: assign random pointers for the copy nodes.
	iter = head;
	while (iter != null) {
		if (iter.random != null) {
			iter.next.random = iter.random.next;
		}
		iter = iter.next.next;
	}

	// Third round: restore the original list, and extract the copy list.
	iter = head;
	RandomListNode pseudoHead = new RandomListNode(0);
	RandomListNode copy, copyIter = pseudoHead;

	while (iter != null) {
		next = iter.next.next;

		// extract the copy
		copy = iter.next;
		copyIter.next = copy;
		copyIter = copy;

		// restore the original list
		iter.next = next;

		iter = next;
	}

	return pseudoHead.next;
}