LeetCode 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.

    /**
     * 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。