zoukankan      html  css  js  c++  java
  • (LinkedList)Intersection of Two Linked Lists

    Write a program to find the node at which the intersection of two singly linked lists begins.

    For example, the following two linked lists:

    A:          a1 → a2
                       ↘
                         c1 → c2 → c3
                       ↗            
    B:     b1 → b2 → b3
    

    begin to intersect at node c1/**

     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) {
     *         val = x;
     *         next = null;
     *     }
     * }
     */
    public class Solution {
        public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
           if(headA==null || headB==null) return null;            //我用的是判断环的方法,碰撞点开始到连接点的长度=head到连接点的长度。
    //一种更好的方法是线计算出2个链表的长度再移动节点。 ListNode moveA=headA; ListNode moveB=headB; ListNode lastA=null; while(moveA!=null && moveA.next!=null) moveA=moveA.next; while(moveB!=null && moveB.next!=null) moveB=moveB.next; if(moveA!=moveB) return null; else { lastA=moveA; moveA.next=headB; } ListNode slow=headA.next; ListNode fast=headA.next.next; while(slow!=fast){ slow=slow.next; fast=fast.next.next; } moveB=slow; moveA=headA; while(moveA!=moveB){ moveA=moveA.next; moveB=moveB.next; } lastA.next=null; return moveB; } }

      

    Notes:

      • If the two linked lists have no intersection at all, return null.
      • The linked lists must retain their original structure after the function returns.
      • You may assume there are no cycles anywhere in the entire linked structure.
      • Your code should preferably run in O(n) time and use only O(1) memory.
  • 相关阅读:
    最近ACM刷题见到的没见过的名词
    最近刷题常常遇见的需要百度的知识点
    简单RPG场景实现(改
    简单RPG场景实现
    位图载入与位图移动
    动态菜单
    静态菜单
    双人五子棋
    HDU 2112 HDU Today
    音痴又音痴的LT
  • 原文地址:https://www.cnblogs.com/kydnn/p/5359439.html
Copyright © 2011-2022 走看看