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.
题目标签：Linked List
题目给了我们两个链表，让我们找到它们是否有交点，返回那个交点。
试想一下，如果是两个长度相等的链表，那么我们只需要遍历链表，比较两个点 是否 相等 就可以了。
所以，我们首先要得到两个链表的长度，如果不一样长，那么把长的链表先走，走到 和 另外一个链表一样长度的时候，开始比较两个点 是否 相等来找到交点；如果没有，那么最后就返回null。
如果两个链表有交点的话，那么从交点前一个点开始，两个链表从这里开始，之后的长度一定是相等的。
Java Solution:
Runtime beats 41.31%
完成日期：06/09/2017
关键词：singly-linked list
关键点：让更长的链表先走完多余的部分
/**
* 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)
{
ListNode cursor1 = headA;
ListNode cursor2 = headB;
int len1 = getListLength(cursor1);
int len2 = getListLength(cursor2); if(len1 > len2)
{
for(int i=0; i<len1-len2;i++)
cursor1 = cursor1.next;
}
else if(len1 < len2)
{
for(int i=0; i<len2-len1;i++)
cursor2 = cursor2.next;
} while(cursor1 != null)
{
if(cursor1 == cursor2)
return cursor1; cursor1 = cursor1.next;
cursor2 = cursor2.next;
} return null;
} private int getListLength(ListNode head)
{
ListNode cursor = head;
int len = 0; while(cursor != null)
{
cursor = cursor.next;
len++;
} return len;
}
}
参考资料：N/A
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