A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

和第133题差不多,都是图的复制,区别在于这道题的label有可能是相同的,所以导致了map的key有可能相同,所以需要处理。

两种方法差不多。第二种更简洁。

1、在复制next之后修改原结构的label为顺序增长,方便建立map,之后再修改回来。

/**
* Definition for singly-linked list with a random pointer.
* class RandomListNode {
* int label;
* RandomListNode next, random;
* RandomListNode(int x) { this.label = x; }
* };
*/
public class Solution {
public RandomListNode copyRandomList(RandomListNode head) {
if( head == null )
return null;
Map map2 = new HashMap<Integer,RandomListNode>();
int num = 1; RandomListNode node = head;
RandomListNode newNode = new RandomListNode(head.label);
RandomListNode node2 = newNode;
map2.put(0,node2);
node.label = 0; node = node.next;
while( node != null ){
RandomListNode nextNode = new RandomListNode(node.label);
node2.next = nextNode;
node2 = node2.next;
node.label = num;
map2.put(num,node2);
num++;
node = node.next; } node = head;
node2 = newNode; while( node != null ){ if( node.random == null){
node = node.next;
node2 = node2.next;
continue;
} node2.random = (RandomListNode) map2.get( node.random.label ); node = node.next;
node2 = node2.next; }
node2 = newNode;
node = head;
while( node != null ){
node.label = node2.label;
node = node.next;
node2 = node2.next;
} return newNode; }
}

2、建立map的时候使用

 Map<RandomListNode,RandomListNode>
/**
* Definition for singly-linked list with a random pointer.
* class RandomListNode {
* int label;
* RandomListNode next, random;
* RandomListNode(int x) { this.label = x; }
* };
*/
public class Solution {
public RandomListNode copyRandomList(RandomListNode head) { if( head == null )
return null;
Map<RandomListNode,RandomListNode> map = new HashMap<RandomListNode,RandomListNode>(); RandomListNode node = head;
while( node != null ){
map.put(node,new RandomListNode(node.label));
node = node.next;
}
node = head;
while( node != null ){ map.get(node).next = map.get(node.next);
map.get(node).random = map.get(node.random);
node = node.next; }
return map.get(head); }
}

leetcode 138. Copy List with Random Pointer ----- java的更多相关文章

  1. Java for LeetCode 138 Copy List with Random Pointer

    A linked list is given such that each node contains an additional random pointer which could point t ...

  2. [LeetCode] 138. Copy List with Random Pointer 拷贝带有随机指针的链表

    A linked list is given such that each node contains an additional random pointer which could point t ...

  3. [LeetCode] 138. Copy List with Random Pointer 拷贝带随机指针的链表

    A linked list is given such that each node contains an additional random pointer which could point t ...

  4. leetcode 138. Copy List with Random Pointer复杂链表的复制

    python代码如下: # Definition for singly-linked list with a random pointer. # class RandomListNode(object ...

  5. Leetcode#138 Copy List with Random Pointer

    原题地址 非常巧妙的方法,不需要用map,只需要O(1)的额外存储空间,分为3步: 1. 先复制链表,但是这个复制比较特殊,每个新复制的节点添加在原节点的后面,相当于"加塞"2. ...

  6. [leetcode]138. Copy List with Random Pointer复制带有随机指针的链表

    public RandomListNode copyRandomList(RandomListNode head) { /* 深复制,就是不能只是复制原链表变量,而是做一个和原来链表一模一样的新链表, ...

  7. 133. Clone Graph 138. Copy List with Random Pointer 拷贝图和链表

    133. Clone Graph Clone an undirected graph. Each node in the graph contains a label and a list of it ...

  8. [Leetcode Week17]Copy List with Random Pointer

    Copy List with Random Pointer 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/copy-list-with-random- ...

  9. 【LeetCode】138. Copy List with Random Pointer

    题目: A linked list is given such that each node contains an additional random pointer which could poi ...

随机推荐

  1. 线程系列3---ThreadLocal类研究

    2013-12-23 17:44:44 Java为线程安全提供了一些工具类,如ThreadLocal类,它代表一个线程局部变量,通过把数据放在ThreadLocal中就可以让每个线程创建一个该变量的副 ...

  2. 【C语言学习】-03 循环结构

    本文目录 循环结构的特点 while循环 do...while循环 for循环 回到顶部 一.循环结构的特点 程序的三种结构: 顺序结构:顺序执行语句 分支结构:通过进行一个判断在两个可选的语句序列之 ...

  3. bzoj 2820 YY的GCD 莫比乌斯反演

    题目大意: 给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对 这里就抄一下别人的推断过程了 后面这个g(x) 算的方法就是在线性 ...

  4. AIX查看内存卡槽

    1.lscfg -vp|grep Processor 2.lscfg -vp|grep -p Memory

  5. vs 折叠跟展开所有方法。

    Ctrl + M + O: 折叠所有方法 Ctrl + M + M: 折叠或者展开当前方法 Ctrl + M + L: 展开所有方法

  6. Cisco IOS Debug Command Reference Command E through H

    debug eap through debug he-module subslot periodic debug eap : to display information about Extensib ...

  7. DotNetBar v12.6.0.4 Fully Cracked

    更新信息: http://www.devcomponents.com/customeronly/releasenotes.asp?p=dnbwf&v=12.6.0.4 如果遇到破解问题可以与我 ...

  8. Java中方法与数组

    1:方法(掌握) (1)方法:就是完成特定功能的代码块. 注意:在很多语言里面有函数的定义,而在Java中,函数被称为方法. (2)格式: 修饰符 返回值类型 方法名(参数类型 参数名1,参数类型 参 ...

  9. 提示错误#165 too few argument in function call

    调用函数时,参数个数少于函数定义.检查一下函数定义和参数调用,两个要一致.

  10. <button>使用注意问题

    最近在项目的上传功能下(IE8)发现了如下的错误: 2015-08-13 09:14:03,396 WARN   [WARN] [http-8080-5] : Handler execution re ...