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;
        }
    }
};