LeetCode上面的题目例如以下：
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.
我的解决方法例如以下:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int lengthA = 0,lengthB = 0;
for(ListNode head= headA;head!=null;head = head.next)
{
lengthA++;
}
for(ListNode head = headB;head!=null;head = head.next)
{
lengthB++;
}
if(lengthA>=lengthB)
{
for(int i=0;i<lengthA-lengthB;i++)
{
headA = headA.next;
}
}
else
{
for(int i=0;i<lengthB-lengthA;i++)
{
headB = headB.next;
}
}
for(ListNode newA =headA,newB = headB;newA!=null&&newB!=null;newA = newA.next,newB = newB.next)
{
if(newA==newB)
{
return newA;
}
}
return null;
}
}
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int lengthA = 0,lengthB = 0;
for(ListNode head= headA;head!=null;head = head.next)
{
lengthA++;
}
for(ListNode head = headB;head!=null;head = head.next)
{
lengthB++;
}
if(lengthA>=lengthB)
{
for(int i=0;i<lengthA-lengthB;i++)
{
headA = headA.next;
}
}
else
{
for(int i=0;i<lengthB-lengthA;i++)
{
headB = headB.next;
}
}
for(ListNode newA =headA,newB = headB;newA!=null&&newB!=null;newA = newA.next,newB = newB.next)
{
if(newA==newB)
{
return newA;
}
}
return null;
}
}