zoukankan      html  css  js  c++  java
  • Intersection of Two Linked Lists(LIST-2 POINTER)

    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.

    FIRST TRY

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
            if(!headA || !headB) return NULL;
            int lenA = 0;
            int lenB = 0;
            ListNode* pA = headA;
            ListNode* pB = headB;
            while(pA->next)
            {
                lenA++;
                pA = pA->next; 
            }
            while(pB->next)
            {
                lenB++;
                pB = pB->next; 
            }
            if(pA == pB)
            {
                pA = headA;
                pB = headB;
                if(lenA > lenB)
                {
                    lenA -= lenB;
                    while(lenA-- > 0) pA = pA->next;
                }
                else if(lenA < lenB)
                {
                    lenB -= lenA;
                    while(lenB-- > 0) pB = pB->next;
                }
                while(pA!=pB)
                {
                    pA = pA->next;
                    pB = pB->next;
                }
                return pA;
            }
            else return NULL;
        }
    };

    Result: Accepted

  • 相关阅读:
    Huffman树与编码
    Python引用复制,参数传递,弱引用与垃圾回收
    Git使用说明
    numpy使用指南
    Python Socket
    温故知新之 数据库的事务、隔离级别、锁
    Oracle数据库的语句级读一致性
    VirtualBox NAT方式与主机互相通信
    Linux的定时任务
    Redis学习
  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/4229839.html
Copyright © 2011-2022 走看看