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。