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.

    /** * 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) {        ListNode *a = headA;        ListNode *b = headB;        if(headA == NULL || headB == NULL)            return NULL;                 int lenA = 0;         int lenB = 0;         while(a && ++lenA){a = a->next;}         while(b && ++lenB){b = b->next;}         a = headA;         b = headB;         if(lenA>lenB)         {             for(int i=0;i<lenA-lenB;i++)             {                 a = a->next;             }         }         else         {             for(int i=0;i<lenB-lenA;i++)             {                 b = b->next;             }         }         while(a && b)         {              if(a==b)                  return a;                            a=a->next;              b=b->next;         }         return NULL;    }};

    解题思路：由于链表a,b有公共序列，且公共序列一定在最后面，不可能出现如下情况

    A:          a1 → a2             a3                   ↘            ↑                     c1 → c2 → c3                   ↗            ↓        B:     b1 → b2 → b3             b4

  • 所以先求2个链表的长度，并将较长的链表的前端先截去，在开始比较指针的地址是否一致。

LeetCode Intersection of Two Linked Lists