Leetcode 160. Intersection of Two Linked Lists/CC150 2.7

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https://leetcode.com/problems/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 遍历linkedList，得到length和tail；

2 Compare tails（reference 对比，不是value对比），如果不同，直接返回；没有intersection；

3 指针指向每个linked List的开始；

4 较长的linked List，移动两个linked list的长度差；

5 遍历每个linked list, 直到指针相同；

C#代码

ListNode FindIntersection(ListNode list1,ListNode list2)        {            if (list1 == null || list2 == null) return null;            //获取tail和size            Result result1 = GetTailAndSize(list1);            Result result2 = GetTailAndSize(list2);            //如果tail reference不同，就没有intersection            if (result1.tail != result2.tail)                return null;            //定位2个指针到linked list的头部            ListNode shorter = result1.size < result2.size ? list1 : list2;            ListNode longer =  result1.size < result2.size ? list2 : list1;            //长linked list的指针向前移动长度差            longer = GetKthNode(longer, Math.Abs(result1.size-result2.size));            //移动两根指针，直到collision            while (shorter != longer)            {                shorter = shorter.next;                longer = longer.next;            }            //任意返回一个            return longer;        }        Result GetTailAndSize(ListNode list)        {            if (list == null)                return null;            int size = 1;            ListNode current = list;            while (current.next != null)            {                size++;                current = current.next;            }            return new Result(current,size);        }        ListNode GetKthNode(ListNode head,int k)        {            ListNode current = head;            while ( k>0 && current != null )            {                current = current.next;                k--;            }            return current;        }

 public class ListNode    {      public int val;      public ListNode next;      public ListNode(int x) { val = x; }  }    public class Result    {        public ListNode tail;        public int size;        public Result(ListNode _tail,int _size)        {            tail = _tail;            size = _size;        }    }

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