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

思路：1. 记录两个list的长度，并检测最后一个元素是否相同。如果不同，表明No intersection。 2. 计算list的长度差offset，让长的list的指针先移动offset，然后两个list的指针一起移动，第一个相同的element就是所求node

2.第二种方法，也很diao！ 代码十分简洁

链接 https://leetcode.com/discuss/17278/accepted-shortest-explaining-algorithm-comments-improvements
3.第二种是先计算listB中value的总和，再将listA中的所有value增加m，再次计算listB中value总和，如果相同no intersection，不同就有！方法仅仅局限于integer

/**
* 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 || !headB) return NULL;
if(headA==headB) return headA;
ListNode *pa=headA;
ListNode *pb=headB;
int la=1;
int lb=1;
while(pa && pa->next) {pa=pa->next;la++;}
while(pb && pb->next) {pb=pb->next;lb++;}
if(pa!=pb) return NULL;
int offset = abs(la-lb);
pa=headA;
pb=headB;
while(offset>0){
if(la>lb) pa=pa->next;
else pb=pb->next;
offset--;
}
while(pa && pb){
if(pa==pb) return pa;
pa=pa->next;
pb=pb->next;
}
}
};