[leetcode] 160.Intersection of Two Linked Lists

题目：

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.

题意：

两个链表，查找是否交叉，如果交叉则找出交叉点。需要O(1)的空间复杂度，O（n）的时间复杂度。

思路：

若不考虑空间复杂度O（1），那么可以使用两个栈从链表头一直装到链表尾，出栈时就是从尾部往头部走。最后一个共同元素就是它们的交叉点。

但是考虑使用O（1）的空间复杂度，那么就不能使用栈。但是考虑到一点，如果有交叉点，那么第一个链表到交叉点有n步，第二个从头到交叉点有m步，那么让其中一个先走｜n－m｜步，那么两者可同时到达交叉点。

以上。

代码如下：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(headA == NULL || headB == NULL)return NULL;
int count1 = 0;
int count2 = 0;
ListNode *t1 = headA, *t2 = headB;
while(t1 !=NULL)
{
count1++;
t1 = t1->next;
}
while(t2 != NULL)
{
count2++;
t2 = t2->next;
}
t1 = headA;
t2 = headB;
if(count2 > count1)
{
for(int i = 0; i < count2 - count1; i++)
t2 = t2->next;
}
else if(count1 > count2)
{
for(int i = 0; i < count1 - count2; i++)
t1 = t1->next;
}
while(t1 != NULL && t2 != NULL && t1 != t2)
{
t1 = t1->next;
t2 = t2->next;
}
return (t1 == t2)?t1:NULL;
}
};