标签：following structure function anywhere original
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.
问题分析，看到该问题，直接想到使用hash函数来处理两个链表，则时间复杂度o(n)空间复杂度o(n)。
然而题目要求，时间复杂度o(n)空间复杂度o(1),因此不能使用hash来处理该问题。我们想两个链表有交集，即后半部分是重合的，两条链表有长有短，重合部分一定存在于最短的链表内。因此，以最短的链表长度，裁剪掉长链表前边的部分，然后和短链表，依次比对链表中的节点，看是否有重合。
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int La=0;
int Lb=0;
ListNode a=headA;
while(a!=null)
{
La++;
a=a.next;
}
ListNode b=headB;
while(b!=null)
{
Lb++;
b=b.next;
}
if(La==Lb)
{
a=headA;
b=headB;
while(a!=null)
{
if(a==b)
return a;
a=a.next;
b=b.next;
}
}
else if(La>Lb)
{
a=headA;
b=headB;
int i=0;
while(i<La-Lb)
{
i++;
a=a.next;
}
}
else
{
a=headA;
b=headB;
int i=0;
while(i<Lb-La)
{
i++;
b=b.next;
}
}
while(a!=null)
{
if(a==b)
return a;
a=a.next;
b=b.next;
}
return null;
}
}
Leetcode#160Intersection of Two Linked Lists
标签：following structure function anywhere original
原文地址：http://7061299.blog.51cto.com/7051299/1640371
