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复杂链表的复制

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)

先按照next遍历复制链表,再对复制后的radom指针赋值。

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public class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;

RandomListNode(int label) {
this.label = label;
}
}

public RandomListNode Clone(RandomListNode pHead) {
if (pHead == null) {
return null;
}
RandomListNode cloneHead = new RandomListNode(pHead.label);
RandomListNode pre = cloneHead;
RandomListNode p = pHead.next;
while (p != null) {
pre.next = new RandomListNode(p.label);
pre = pre.next;
p = p.next;
}
for (RandomListNode pt = pHead, cp = cloneHead; pt != null; pt = pt.next, cp = cp.next) {
if (pt.random != null) {
cp.random = findNode(pt.random.label, cloneHead);
}
}
return cloneHead;
}

public RandomListNode findNode(int label, RandomListNode head) {
for (RandomListNode p = head; p != null; p = p.next) {
if (p.label == label) {
return p;
}
}
return null;
}

网上题解

链接:https://www.nowcoder.com/questionTerminal/f836b2c43afc4b35ad6adc41ec941dba?answerType=1&f=discussion
来源:牛客网

不用开辟新的Map,但是其实需要多次遍历。
图片说明

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/*
*解题思路:
*1、遍历链表,复制每个结点,如复制结点A得到A1,将结点A1插到结点A后面;
*2、重新遍历链表,复制老结点的随机指针给新结点,如A1.random = A.random.next;
*3、拆分链表,将链表拆分为原链表和复制后的链表
*/
public class Solution {
public RandomListNode Clone(RandomListNode pHead) {
if(pHead == null) {
return null;
}

RandomListNode currentNode = pHead;
//1、复制每个结点,如复制结点A得到A1,将结点A1插到结点A后面;
while(currentNode != null){
RandomListNode cloneNode = new RandomListNode(currentNode.label);
RandomListNode nextNode = currentNode.next;
currentNode.next = cloneNode;
cloneNode.next = nextNode;
currentNode = nextNode;
}

currentNode = pHead;
//2、重新遍历链表,复制老结点的随机指针给新结点,如A1.random = A.random.next;
while(currentNode != null) {
currentNode.next.random = currentNode.random==null?null:currentNode.random.next;
currentNode = currentNode.next.next;
}

//3、拆分链表,将链表拆分为原链表和复制后的链表
currentNode = pHead;
RandomListNode pCloneHead = pHead.next;
while(currentNode != null) {
RandomListNode cloneNode = currentNode.next;
currentNode.next = cloneNode.next;
cloneNode.next = cloneNode.next==null?null:cloneNode.next.next;
currentNode = currentNode.next;
}

return pCloneHead;
}
}