题目描述
输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针random指向一个随机节点），请对此链表进行深拷贝，并返回拷贝后的头结点。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）
思路以及解法
一共有三种写法，第一种就是下面的，先复制next链路上的节点。然后复制random节点。复制random节点的时候，采取的是遍历的做法。
class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}
public class Solution {
public static void main(String[] args) {
RandomListNode node1 = new RandomListNode(1);
RandomListNode node2 = new RandomListNode(2);
RandomListNode node3 = new RandomListNode(3);
RandomListNode node4 = new RandomListNode(4);
RandomListNode node5 = new RandomListNode(5);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;
node2.random = node1;
node1.random = node3;
Solution solution = new Solution();
RandomListNode newNode = solution.Clone(node1);
}
public RandomListNode Clone(RandomListNode pHead) {
if (pHead == null) {
return null;
}
RandomListNode head = new RandomListNode(pHead.label);
RandomListNode newHead = head;
RandomListNode oldHead = pHead;
while (oldHead.next != null) {
RandomListNode next = new RandomListNode(oldHead.next.label);
newHead.next = next;
oldHead = oldHead.next;
newHead = newHead.next;
}
RandomListNode newRandomHead = head;
RandomListNode oldRandomHead = pHead;
while (oldRandomHead != null) {
// 找到第一个节点的随机节点
RandomListNode randomNode = oldRandomHead.random;
if (randomNode != null) {
RandomListNode temp = pHead;
RandomListNode tempNew = head;
while (temp != randomNode) {
temp = temp.next;
tempNew = tempNew.next;
}
newRandomHead.random = tempNew;
}
oldRandomHead = oldRandomHead.next;
newRandomHead = newRandomHead.next;
}
return head;
}
}
第二种就是上面写法的优化版本，因为我们在复制random的时候，花费了太多时间在查找节点的上面，优化就是用空间换时间，用hashmap存储旧节点和新节点的映射，就可以直接查找对应的random节点。
import java.util.HashMap;
class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}
public class Solution {
public static void main(String[] args) {
RandomListNode node1 = new RandomListNode(1);
RandomListNode node2 = new RandomListNode(2);
RandomListNode node3 = new RandomListNode(3);
RandomListNode node4 = new RandomListNode(4);
RandomListNode node5 = new RandomListNode(5);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;
node2.random = node1;
node1.random = node3;
Solution solution = new Solution();
RandomListNode newNode = solution.Clone(node1);
System.out.println(newNode);
}
public RandomListNode Clone(RandomListNode pHead) {
if (pHead == null) {
return null;
}
RandomListNode head = new RandomListNode(pHead.label);
RandomListNode newHead = head;
RandomListNode oldHead = pHead;
HashMap<RandomListNode,RandomListNode>map =new HashMap<>();
map.put(oldHead,newHead);
while (oldHead.next != null) {
RandomListNode next = new RandomListNode(oldHead.next.label);
newHead.next = next;
oldHead = oldHead.next;
newHead = newHead.next;
map.put(oldHead,newHead);
}
RandomListNode newRandomHead = head;
RandomListNode oldRandomHead = pHead;
while (oldRandomHead != null) {
// 找到第一个节点的随机节点
RandomListNode randomNode = oldRandomHead.random;
if (randomNode != null) {
newRandomHead.random = map.get(randomNode);
}
oldRandomHead = oldRandomHead.next;
newRandomHead = newRandomHead.next;
}
return head;
}
}
第三种方法就是不需要map，直接在之前的链表上复制，每一个节点都复制一个节点跟在后面，之后再复制random节点。都复制完成之后，将复制的新节点连接起来。
public RandomListNode Clone(RandomListNode pHead) {
if (pHead == null) {
return null;
}
RandomListNode oldHead = pHead;
while (oldHead != null) {
RandomListNode head = new RandomListNode(oldHead.label);
RandomListNode temp = oldHead.next;
oldHead.next = head;
head.next = temp;
oldHead = temp;
}
RandomListNode randomHead = pHead;
while (randomHead != null) {
RandomListNode random = randomHead.random;
if (random != null) {
randomHead.next.random = random.next;
}
randomHead = randomHead.next.next;
}
RandomListNode pCloneHead = pHead.next;
RandomListNode cur = pHead;
while (cur.next != null) {
RandomListNode next = cur.next;
cur.next = next.next;
cur = next;
}
return pCloneHead;
}
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