[LeetCode160]Intersection of Two Linked Lists

题目：

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：求一个单链表交集的首结点

　　方法一：找出两个链表的长度差n，长链表先走n步，然后同时移动，判断有没有相同结点

　　方法二：两个链表同时移动，链表到尾部后，跳到链表头部，相遇点即为所求

代码：

方法一：

 /**
 * 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) {
         //方法一
         ListNode *p = headA;
         ListNode *q = headB;
         if(!p || !q) return NULL;
         while(p && q && p != q)
         {
         p = p->next;
         q = q->next;
         if(p == q)
         return p;
         if(!p)
         p = headB;
         if(!q)
         q = headA;
         }
         return p;
     }
 };

方法二：

 /**
 * 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) {
         //方法二
         if (headA == NULL || headB == NULL) return NULL;
         ListNode *p = headA;
         ListNode *q = headB;
         int countA = , countB = ;
         while (p->next)
         {
             p = p->next;
             ++countA;
         }
         while (q->next)
         {
             q = q->next;
             ++countB;
         }
         if (p != q) return NULL;
         if (countA > countB)
         {
             for (int i = ; i < countA - countB; ++i)
                 headA = headA->next;
         }
         else if (countB > countA)
         {
             for (int i = ; i < countB - countA; ++i)
                 headB = headB->next;
         }
         while (headA != headB)
         {
             headA = headA->next;
             headB = headB->next;
         }
         return headA;
 
     }
 };