zoukankan      html  css  js  c++  java
  • Intersection of Two Linked Lists 解答

    Question

    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.

    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.

    Solution

    Key to the solution here is to traverse two lists to get their lengths. Therefore, we can move the pointer for the longer list first and then compare elements of both lists.

     1 /**
     2  * Definition for singly-linked list.
     3  * public class ListNode {
     4  *     int val;
     5  *     ListNode next;
     6  *     ListNode(int x) {
     7  *         val = x;
     8  *         next = null;
     9  *     }
    10  * }
    11  */
    12 public class Solution {
    13     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    14         if (headA == null || headB == null)
    15             return null;
    16         ListNode p1 = headA, p2 = headB;
    17         int l1 = 0, l2 = 0, ll = 0;
    18         while (p1 != null) {
    19             l1++;
    20             p1 = p1.next;
    21         }
    22         while (p2 != null) {
    23             l2++;
    24             p2 = p2.next;
    25         }
    26         p1 = headA;
    27         p2 = headB;
    28         if (l2 >= l1) {
    29             ll = l2 - l1;
    30             while (ll > 0) {
    31                 p2 = p2.next;
    32                 ll--;
    33             }
    34         } else {
    35             ll = l1 - l2;
    36             while (ll > 0) {
    37                 p1 = p1.next;
    38                 ll--;
    39             }
    40         }
    41         while (p1 != null && p2 != null) {
    42             if (p1.val == p2.val)
    43                 return p1;
    44             p1 = p1.next;
    45             p2 = p2.next;
    46         }
    47         return null;
    48     }
    49 }
  • 相关阅读:
    什么是wsgi,uwsgi,uWSGI
    Flask 和 Django 路由映射的区别
    简述浏览器通过WSGI请求动态资源的过程
    前端qq交流群
    python qq交流群
    python 魔法方法 __str__和__repr__
    python 使用for 实现死循环
    查看Django版本
    pep8 python 编码规范
    python random.randint(9,10)结果是什么?
  • 原文地址:https://www.cnblogs.com/ireneyanglan/p/4812526.html
Copyright © 2011-2022 走看看