leetcode 160

160. 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.

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.

找到两个单链表相交的点。此处假设单链表不存在环。

思路：首先获得两个单链表的长度，得到长度的差值n，然后将指向长链表的指针A先向后移动n位，之后指针A同指向短链表的指针B同时每次移动一位，当A与B指向同一节点时，该节点就是相交的结点。

 /**
  * 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 == NULL || headB == NULL)
         {
             return NULL;
         }
         ListNode * a = headA;
         ListNode * b = headB;
         int n = ;
         int m = ;
         while(headA->next != NULL)
         {
             headA = headA->next;
             n++;
         }
         while(headB->next != NULL)
         {
             headB = headB->next;
             m++;
         }
         if(headA != headB)
         {
             return NULL;
         }
         if(n > m)
         {
             for(int i = ; i < n-m; i++)
             {
                 a = a->next;
             }
             while(a)
             {
                 if(a == b)
                 {
                     return a;
                 }
                 else
                 {
                     a = a->next;
                     b = b->next;
                 }
             }
         }
         else
         {
             for(int i = ; i < m-n; i++)
             {
                 b = b->next;
             }
             while(b)
             {
                 if(a == b)
                 {
                     return a;
                 }
                 else
                 {
                     a = a->next;
                     b = b->next;
                 }
             }
         }
         return NULL;
     }
 };

拓展：

http://www.cppblog.com/humanchao/archive/2015/03/22/47357.html