翻译

写一个程序来找出两个单链表相交的开始处。

例如，如下有两个链表：

在节点c1处开始相交。

批注：
如果两个链表根本没有相交，返回NULL。
在函数返回后链表必须保留原有的数据结构。
你可以假设在整个链结构中没有循环。
你的代码最后可以在O(n)时间和O(1)空间运行。

原文

Write a program to find the node at which the intersection of two singly linked lists begins.


For example, the following two linked lists:

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.

分析

注意这里的情况是节点相交，而不仅仅是那些值相等而已。

看我图片画的这么认真，还不关注我博客嘛 ^_^

代码

/**
* 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;
        ListNode *listA = headA, *listB = headB;
        while (listA && listB) {
            if (listA == listB) return listA;
            listA = listA->next;
            listB = listB->next;

            if (listA == listB) return listA;
            if (listA == NULL) listA = headB;
            if (listB == NULL) listB = headA;
        }
        return listA;
    }
};

updated at 2016/09/20

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        ListNode listA = headA, listB = headB;
        while (listA != null && listB != null) {
            if (listA == listB) return listA;
            listA = listA.next;
            listB = listB.next;
            if (listA == listB) return listA;
            if (listA == null) listA = headB;
            if (listB == null) listB = headA;
        }
        return listA;
    }
}