[leetcode] 160. Intersection of Two Linked Lists 解题报告

题目链接：https://leetcode.com/problems/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.

思路：先计算两个链表的长度，让长的先走他们长度的差值，然后这时候他们剩余长度相等了，再一起走，直到找到公共的结点，或者直到结束。恩，思路还是比较清晰的！Have a nice day！

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 || !headB)  return NULL;
int len1 = 0, len2 = 0;
ListNode *p= headA, *q = headB;
while(p)
{
len1++;
p = p->next;
}
while(q)
{
len2++;
q = q->next;
}
p = headA, q = headB;
int i = 0;
while(i < abs(len1 - len2))
{
if(len1 > len2)
p = p->next;
else
q = q->next;
i++;
}
while(p && q)
{
if(p != q)
{
p = p->next;
q = q->next;
}
else
return p;
}
return NULL;
}
};

哦，看了下discus，发现一个极其简约的代码，不禁感叹想出这个方式的人好聪明！这种算法的思想是两个指针各走len1 + len2的长度，最后将会同步，然后判断是否相等，其时间复杂度是O(len1 + len2)

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 || !headB)  return NULL;
ListNode *p= headA, *q = headB;
while(p != q)
{
p = p==NULL?headB:p->next;
q = q==NULL?headA:q->next;
}
return p;
}
};

参考：https://leetcode.com/discuss/66203/java-solution-without-knowing-the-difference-in-len