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Leetcode 160:Intersection of Two Linked Lists

时间:2016-02-26 00:37:13      阅读:187      评论:0      收藏:0      [点我收藏+]

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.

class Solution {
public:
   ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
   {
    ListNode *p1 = headA;
    ListNode *p2 = headB;

    if (p1 == NULL || p2 == NULL) return NULL;

    while (p1 != NULL && p2 != NULL && p1 != p2) 
    {
        p1 = p1->next;
        p2 = p2->next;

        if (p1 == p2) return p1; //当达到相遇点时,直接返回p1或者p2

        if (p1 == NULL) p1 = headB; //当p1比p2先到达结尾时,重新将其移动到第二个链表的头部
        if (p2 == NULL) p2 = headA; //当p2比p1先到达结尾时,重新将其移动到第二个链表的头部,这样的话,二者到相遇点的距离是相等的
    }

    return p1;
   }
};

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Leetcode 160:Intersection of Two Linked Lists

原文:http://blog.csdn.net/geekmanong/article/details/50743577

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