所谓带随机指针的链表,结构如下:
class Node {
int val;
Node next;
Node random;
public Node(int val) {
this.val = val;
this.next = null;
this.random = null;
}
}
除next外,还有一个随机指针random,随机指向链表中的某个元素(当然 :random也可能为null). 复制的难度在于, 新节点刚new出来时,其random指向的另外1个“新”节点,可能还没复制出来(即:首次无法确定新节点的random该指向谁,除非所有老节点全复制完)
有二种做法:
1、借助额外的Map记录“新-老”节点的映射
public Node copyRandomList(Node head) {
if (head==null){
return null;
}
Node h = head;
Map<Node,Node> map = new HashMap<>();
Node newHead = new Node(head.val);
Node curr = newHead;
//第一轮,复制节点,random挂空,同时记录处理过的老节点与新节点的映射关系
while(head!=null){
if (head.next!=null){
Node newNext = new Node(head.next.val);
curr.next = newNext;
}
map.put(head,curr);
head = head.next;
curr = curr.next;
}
head = h;
//第二轮,处理random指向,通过map映射,查到random指向的新节点
while(head!=null){
Node newNode = map.get(head);
if (head.random!=null){
newNode.random = map.get(head.random);
}
head = head.next;
}
return newHead;
}
2、如果要求空间复杂度为O(1),还有1个比较巧妙的思路
a、 先把每个节点复制一份, 挂在自己身后,相当于A -> B -> C 变成 A -> A' -> B -> B' -> C -> C'
b、 每2轮,处理random时,通过身后的“影子”,就能找到random的新节点在哪
c、 将链表分离, A -> A' -> B -> B' -> C -> C' 变成 A -> B -> C 和A' -> B' -> C' 返回A'
public Node copyRandomList(Node head) {
if (head==null){
return head;
}
Node curr = head;
//复制自身,挂在身后
while(curr!=null){
Node newNode = new Node(curr.val);
newNode.next = curr.next;
curr.next = newNode;
curr = curr.next.next;
}
//复制random
curr = head;
while(curr!=null){
if (curr.random!=null){
curr.next.random = curr.random.next;
}
curr = curr.next.next;
}
//分离链表
curr = head;
Node newHead = head.next;
while(curr!=null){
Node newCurr = curr.next;
curr.next = curr.next.next;
if (newCurr.next!=null){
newCurr.next = newCurr.next.next;
}
curr = curr.next;
}
return newHead;
}