【题目】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)
【思路】
【实现】
/*
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;
}
//第一步
RandomListNode currentNode = pHead;
while(currentNode != null){
RandomListNode cloneNode = new RandomListNode(currentNode.label);
RandomListNode nextNode = currentNode.next;
currentNode.next = cloneNode;
cloneNode.next = nextNode;
currentNode = nextNode;
}
//第二步
currentNode = pHead;
while(currentNode != null){
currentNode.next.random = currentNode.random == null?null:currentNode.random.next;
currentNode = currentNode.next.next;
}
//第三步
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;
}
}
【代码解析】
第一步:复制每个结点,如复制结点A得到A‘,将结点A’插到结点A后面;
RandomListNode currentNode = pHead;
while(currentNode != null){
RandomListNode cloneNode = new RandomListNode(currentNode.label);
RandomListNode nextNode = currentNode.next;//保存一个节点nextNode
currentNode.next = cloneNode;//将cloneNode连在currentNode后面
cloneNode.next = nextNode;//将保存的nextNode节点连在复制的节点的后面
currentNode = nextNode;
}
第二步: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;
currentNode = pHead;
while(currentNode != null){
currentNode.next.random = currentNode.random == null?null:currentNode.random.next;
currentNode = currentNode.next.next;
}
第三步: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;
}
参考:
1.《剑指offer》