1.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,另一个特殊指针指向任意一个节点),返回结果为复制后复杂链表的head。(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)
2.解题思路
这题我没懂。代码借鉴别人的
3.代码
/*
public class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}
*/
public class Solution {
public RandomListNode Clone(RandomListNode pHead)
{
if (pHead == null) return null;
RandomListNode head = new RandomListNode(pHead.label);
RandomListNode ans = head;
if (pHead.random != null) {
head.random = new RandomListNode(pHead.random.label);
}
while (pHead.next != null) {
pHead = pHead.next;
head.next = new RandomListNode(pHead.label);
if (pHead.random != null) {
head.next.random = new
RandomListNode(pHead.random.label);
}
head = head.next;
}
return ans;
}
}
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;
}
}