【LeetCode & 剑指offer刷题】链表题5：52 两个链表的第一个公共结点（Intersection of Two Linked Lists）
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
52 两个链表的第一个公共结点
题目描述
输入两个链表，找出它们的第一个公共结点。
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
#include
class Solution
{
public :
ListNode * FindFirstCommonNode ( ListNode * pHead1 , ListNode * pHead2 )
{
//求两链表的长度
int len1 = findListLength ( pHead1 );
int len2 = findListLength ( pHead2 );
ListNode * plong = pHead1 , * pshort = pHead2 ;
if ( len1 < len2 )
{
pshort = pHead1 ;
plong = pHead2 ;
}
for ( int i = 1 ; i <= abs ( len1 - len2 ); i ++) plong = plong -> next ; //较长的链表多走几步
//同时步进，直到遇到相同结点或者均遇到尾结点
while ( plong != pshort )
{
plong = plong -> next ;
pshort = pshort -> next ;
}
return plong ;
}
private :
int findListLength ( ListNode * p )
{
int n = 0 ;
while ( p != nullptr )
{
p = p -> next ;
n ++;
}
return n ;
}
};
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.
Credits:
Special thanks to @stellari for adding this problem and creating all test cases.
C++
//问题：求两链表的交汇点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/*
方法：双指针法
(1) 如果有交汇点，p1扫描A，p2扫描B，扫描到结尾后，p1重定向到headB,p2重定向到headA,之后一定会在交汇点处相遇
因为交汇点之后都是路径相同的，交汇点之前的路径差可以由互换的两次扫描中抵消
(2) 如果没有交汇点，p1最后会到b末尾，p2会到a末尾，p1=p2=null，退出程序
O(m+n),O(1)
*/
class Solution
{
public :
ListNode * getIntersectionNode ( ListNode * headA , ListNode * headB )
{
ListNode * p1 = headA ;
ListNode * p2 = headB ;
if ( p1 == NULL || p2 == NULL ) return NULL ;
while ( p1 && p2 && p1 != p2 ) //只要不为空，进行扫描(注意，加上p1!=p2的判断，可能两链表长度为1，且相交，不加的话会返回null)
{
p1 = p1 -> next ;
p2 = p2 -> next ;
if ( p1 == p2 ) return p1 ; //p1,p2同时为nullptr或者指向交汇点
if ( p1 == NULL ) p1 = headB ; //重定向到另一个链表首结点
if ( p2 == NULL ) p2 = headA ;
}
return p1 ;
}
};
posted @ 2019-01-05 16:53 wikiwen 阅读( ...) 评论( ...) 编辑 收藏