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。
我的解答
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/*
public class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}
*/
import java.util.ArrayList;
public class Solution {
public RandomListNode Clone(RandomListNode pHead) {
if (null == pHead) {
return null;
}
ArrayList<RandomListNode> randomNodes = new ArrayList<>();
RandomListNode cloneHead = new RandomListNode(pHead.label);
if (null != pHead.random) {
RandomListNode random = new RandomListNode(pHead.random.label);
randomNodes.add(random);
cloneHead.random = random;
}
RandomListNode prev = cloneHead;
while(null != pHead.next) {
pHead = pHead.next;
RandomListNode random = null;
if (null != pHead.random) {
random = new RandomListNode(pHead.random.label);
randomNodes.add(random);
}
RandomListNode curNode = findInRandomList(randomNodes, pHead);
if (null == curNode) {
curNode = new RandomListNode(pHead.label);
}
prev.next = curNode;
curNode.random = random;
prev = curNode;
}
return cloneHead;
}
private RandomListNode findInRandomList(ArrayList<RandomListNode> randomNodes, RandomListNode node) {
RandomListNode result = null;
for (RandomListNode n : randomNodes) {
if (n.label == node.label) {
result = n;
break;
}
}
if (null != result) {
randomNodes.remove(result);
}
return result;
}
}
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