Intersection of Two Linked Lists 解答

提示：扫一扫查出行
【扫一扫了解最新限行尾号】
复制提示

Question

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.

Solution

Key to the solution here is to traverse two lists to get their lengths. Therefore, we can move the pointer for the longer list first and then compare elements of both lists.

/**
* 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) {
if (headA == null || headB == null)
return null;
ListNode p1 = headA, p2 = headB;
int l1 = 0, l2 = 0, ll = 0;
while (p1 != null) {
l1++;
p1 = p1.next;
}
while (p2 != null) {
l2++;
p2 = p2.next;
}
p1 = headA;
p2 = headB;
if (l2 >= l1) {
ll = l2 - l1;
while (ll > 0) {
p2 = p2.next;
ll--;
}
} else {
ll = l1 - l2;
while (ll > 0) {
p1 = p1.next;
ll--;
}
}
while (p1 != null && p2 != null) {
if (p1.val == p2.val)
return p1;
p1 = p1.next;
p2 = p2.next;
}
return null;
}
}