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.
这个题目是要找出两个链表的交叉点,该如何解决呢?如果两个链表相交,我们从A链表出发移动一段距离alen,出B列表出发移动一段距离blen,那么会发现他们指向同一个节点c1。那么这个距离是多少呢?
我们把每一个链表看成两段,不相交的一段和相交的一段。相交的一段对于两个链表长度是一样的,不想交的一段链表的长度是不同的。如果我们分别计算出两个链表的长度,然后计算他们长度的差值f,然后在较长的链表上先移动距离f,再同时从A,B链表出发开始遍历,若发现他们指向同一个节点,则他们相交并返回相交的点,若他们不想交,则会遍历到链表的尾部,则返回null。代码如下:
/**
* 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) {
ListNode p1 = headA, p2 = headB;
int len1 = 0, len2 = 0;
while (p1 != null) { //求链表A的长度
p1 = p1.next;
len1++;
}
while (p2 != null) { //求链表B的长度
p2 = p2.next;
len2++;
}
p1 = headA;
p2 = headB;
if (len1 > len2) { //计算链表长度的差值并在较长的链表上向后移动|len1-len2|
for (int i = 0;i < len1 - len2; i++) {
p1 = p1.next;
}
} else {
for (int i = 0;i < len2 - len1; i++) {
p2 = p2.next;
}
}
while (p1 != p2) { //向后遍历链表A和链表B,找到相交的节点,若遍历到最后,返回null
p1 = p1.next;
p2 = p2.next;
}
return p1;
}
}