[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 = 1, countB = 1;
        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 = 0; i < countA - countB; ++i)
                headA = headA->next;
        }
        else if (countB > countA)
        {
            for (int i = 0; i < countB - countA; ++i)
                headB = headB->next;
        }
        while (headA != headB)
        {
            headA = headA->next;
            headB = headB->next;
        }
        return headA;

    }
};