码迷,mamicode.com
首页 > 其他好文 > 详细

Intersection of Two Linked Lists

时间:2015-01-06 09:56:09      阅读:174      评论:0      收藏:0      [点我收藏+]

标签:algorithm   leetcode   

题目描述:

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.

思路分析:先求出两段链表的长度length_a、length_b,让较长链表的头指针先移动|length_a-length_b|个单位。然后让两个链表的头指针一起移动,当两个指向两个链表的指针相等时,即找到了交点。


代码:

ListNode * Solution::getIntersectionNode(ListNode *headA, ListNode *headB)
{
    ListNode * a = headA;
    ListNode * b = headB;
    int length_a = 0,length_b = 0;
    while(a != NULL)
    {
        length_a++;
        a = a->next;
    }
    while(b != NULL)
    {
        length_b++;
        b = b->next;
    }
    if(length_a > length_b)
    {
        int difference = length_a - length_b;
        while(difference > 0)
        {
            difference--;
            headA = headA->next;
        }

    }
    else if(length_b > length_a)
    {
        int difference = length_b - length_a;
        while(difference > 0)
        {
            difference--;
            headB = headB->next;
        }
    }
    while(headA != headB)
    {
        headA = headA->next;
        headB = headB->next;
    }
    return headA;
}


Intersection of Two Linked Lists

标签:algorithm   leetcode   

原文地址:http://blog.csdn.net/yao_wust/article/details/42454151

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!