Multi-directional linked list structure such as
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public class Node{ public int value; public Node next; public Node rand; public Node(int data){ this.value = data; } }
Method 1: Use HashMap structure
public class CopyFromMultiNode { public static void main(String[] args){ int[] array = {12,3,4,5,6,77,6,54,56,6,7,87,15,15,15}; //数组转Node Node head = array2node(array); Node help = head; System.out.print("处理前 "); while(help != null){ System.out.print(help.value + " "); help = help.next; } //使用HashMap结构 Node res = copyFromRand1(head); System.out.println(); System.out.print("处理后结果:"); while(res != null){ System.out.print(res.value+" "); res = res.next; } } //使用HashMap结构 public static Node copyFromRand1(Node head){ Node cur = head; HashMap<Node, Node> map = new HashMap<>(); while(cur != null){ map.put(cur, new Node(cur.value)); cur = cur.next; } cur = head; while(cur != null){ map.get(cur).next = map.get(cur.next); map.get(cur).rand = map.get(cur.rand); cur = cur.next; } return map.get(head); } //数组转Node功能,供测试使用 public static Node array2node(int[] array){ Node head = new Node(array[0]); Node cur = head; for(int i=1; i<array.length; i++){ cur.next = new Node(array[i]); cur = cur.next; } return head; } //基础node节点结构 public static class Node{ public int value; public Node next; public Node rand; public Node(int data){ this.value = data; } } }
2. Use several valid variable methods without other structures
//使用几个有效变量方法 //替换方法一的copyFromRand1方法 public static Node copyFromRand2(Node head){ Node next = null; Node cur = head; //1 -> 2 -> 3 -> 4 ==> 1 -> 1` -> 2 -> 2` -> 3 -> 3` -> 4 -> 4 //完成链表拼接 while(cur != null){ next = cur.next; cur.next = new Node(cur.value); cur.next.next = next; cur = next; } cur = head; Node curCopy = null; //添加Node的rand值 while(cur != null){ next = cur.next.next; curCopy = cur.next.next; curCopy = cur.rand != null? cur.rand.next: null; cur = next; } Node res = head.next; cur = head; //拆分 // 1 -> 1` -> 2 -> 2` -> 3 -> 3` -> 4 -> 4 // ==> 1 -> 2 -> 3 -> 4 和 1`-> 2`-> 3`-> 4` while(cur != null){ next = cur.next.next; curCopy = cur.next; cur.next = next; curCopy.next = next != null ? next.next:null; cur = next; } return res; }
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