• 【数据结构】算法 Copy List with Random Pointer 复制带随机指针的链表


    Copy List with Random Pointer 复制带随机指针的链表

    Description

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

    Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

    For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

    Return the head of the copied linked list.

    The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

    • val: an integer representing Node.val
    • random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

    Your code will only be given the head of the original linked list.

    将一个链表,按照给定的x值,分成2部分,链表前部小于x,后部大于等于x

    img

    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]]
    
    

    思路:

    将每个节点都复制一份并插入在原始节点之后,所有random指向原始节点,将复制出的节点中的random指向原始节点的后一个。最后将复制的节点两两拆开。

     /*
    // 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 null;
            }
             
            Node p = head;
            Node q = null;
            while (p!=null){
                q = new Node(p.val);             
                q.random = p.random;
                q.next = p.next;
                p.next = q;
                p = q.next;
            }
            p = head.next;
            while(p!=null){
                if (p.random!=null){
                    p.random = p.random.next;
                }
                else{
                    p.random = null;
                }
    
               p = p.next;
                if (p!=null){
                    p = p.next;
                }
            }
            Node newhead = head.next;
            p = head;
            while(p!=null){
                q = p.next;
                p.next = q.next;
                if (p.next!=null){
                    q.next = p.next.next;
                }
                p = p.next;
            }
            return newhead;
        }
    }
    
  • 相关阅读:
    仿照Excel的控件,支持c#
    Excel地图插件
    com接口调用
    决策树算法
    python练习题
    numpy练习题
    机器学习_线性回归
    python学习之老男孩python全栈第九期_数据库day005知识点总结 —— MySQL数据库day5
    problem-solving-with-algorithms-and-data-structure-usingpython(使用python解决算法和数据结构) -- 基本数据结构 -- 队列
    python学习之老男孩python全栈第九期_数据库day004知识点总结 —— MySQL数据库day4
  • 原文地址:https://www.cnblogs.com/dreamtaker/p/14539925.html
Copyright © 2020-2023  润新知