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.

双指针法：

定义两个指针，分别指向 headA和headB，分别求出两的链结点个数（长度），然后  求出两者差r，让长的链表 先走r步，然后从长的链的当前位置和短链的头部一起往后遍历，知道找到相同的,则为所求

/** * 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 *pa=headA,*pb=headB;        if(headA==NULL||headB==NULL)return NULL;  //有一个为空则返回为空。   int m=0,n=0,r;   while(pa){m++;pa=pa->next;}     while(pb){n++,pb=pb->next;}   pa=headA; //pa重新回到头节点位置   pb=headB;   if(m>n){       r=m-n;   while(r){pa=pa->next;r--;}   }    else{        r=n-m;   while(r){pb=pb->next;r--;}    }   while(pa!=pb){              pa=pa->next;       pb=pb->next;   }   return pa;     }       };