160. 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.
题目大意：
找出两个链表后半部分的交汇点。
思路：
1.求出两个链表的长度。
2.获取链表长度差n。
3.将长的链表先移动到第n个节点。
4.对长链表和短链表进行比较。(同时向后移动)如果在链表尾之前找到相等的节点，返回该节点，如果没找到，返回NULL。
代码如下：
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class
Solution {
public
:
int
listLength(ListNode *head)
//用快指针求链表长度
{
ListNode * p = head;
int
i = 0 ;
while
(p && p->next)
{
i++;
p = p->next->next;
}
if
(p == NULL)
return
2 * i;
return
2 * i + 1;
}
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
int
lenA = listLength(headA);
int
lenB = listLength(headB);
int
maxLen = lenA > lenB ? lenA :lenB;
int
remain ;
ListNode * la,*lb;
la = headA;
lb = headB;
if
(maxLen == lenA)
{
remain = lenA - lenB;
while
(remain--)
{
la = la->next;
}
}
else
{
remain = lenB - lenA;
while
(remain--)
lb = lb->next;
}
while
(lb != NULL)
{
if
(la != lb)
{
la = la->next;
lb = lb->next;
}
else
{
return
la;
}
}
return
NULL;
}
};
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本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1837480
