请实现 copyRandomList 函数,复制一个复杂链表。在复杂链表中,每个节点除了有一个 next 指针指向下一个节点,还有一个 random 指针指向链表中的任意节点或者 null。
示例 1:
输入:head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
输出:[[7,null],[13,0],[11,4],[10,2],[1,0]]
示例 2:
输入:head = [[1,1],[2,1]]
输出:[[1,1],[2,1]]
示例 3:
输入:head = [[3,null],[3,0],[3,null]]
输出:[[3,null],[3,0],[3,null]]
示例 4:
输入:head = []
输出:[]
解释:给定的链表为空(空指针),因此返回 null。
提示:
-10000 <= Node.val <= 10000
Node.random 为空(null)或指向链表中的节点。
节点数目不超过 1000 。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
/*
// Definition for a Node.
class Node {
int val;
Node next;
Node random;
public Node(int val) {
this.val = val;
this.next = null;
this.random = null;
}
}
*/
class Solution {
public Node copyRandomList(Node head) {
if(head == null){
return head;
}
Node curr = head;
while(curr != null){
Node copyNode = new Node(curr.val);
copyNode.next = curr.next;
curr.next = copyNode;
curr = copyNode.next;
}
curr = head;
while(curr != null && curr.next != null){
Node tempNode = curr.next;
if(curr.random != null){
tempNode.random = curr.random.next;
}else{
tempNode.random = null;
}
curr = curr.next.next;
}
Node result = new Node(0);
Node origin = head;
curr = head.next;
result.next = curr;
while(curr != null && curr.next != null){
origin.next = curr.next;
origin = origin.next;
curr.next = curr.next.next;
curr = curr.next;
}
return result.next;
}
}
看了题解,大佬解法,思路清晰。
/*
// Definition for a Node.
class Node {
int val;
Node next;
Node random;
public Node(int val) {
this.val = val;
this.next = null;
this.random = null;
}
}
*/
class Solution {
public Node copyRandomList(Node head) {
if(head == null){
return head;
}
Map map = new HashMap();
Node curr = head;
while(curr != null){
map.put(curr, new Node(curr.val));
curr = curr.next;
}
curr = head;
while(curr != null){
map.get(curr).next = map.get(curr.next);
map.get(curr).random = map.get(curr.random);
curr = curr.next;
}
return map.get(head);
}
}