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.
Special thanks to @stellari for adding this problem and creating all test cases.
---
Solution: Get the lengths of each list. Make them the same length by skipping the first few node of the longer list. For each of the remaining nodes, check each one to see if they are the same node. If not found, then 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) {
// Find the lengths of the two lists.
int lengthA = getLength(headA);
int lengthB = getLength(headB);
// Make both lists the same length by skipping the
// first few nodes of one of the lists if needed.
for (int i = lengthB; i < lengthA; i++) {
headA = headA.next;
}
for (int i = lengthA; i < lengthB; i++) {
headB = headB.next;
}
// Check each node to see if they are the same node.
while (headA != null) {
if (headA == headB) return headA;
headA = headA.next;
headB = headB.next;
}
// No intersection found.
return null;
}
private int getLength(ListNode node) {
int length = 0;
while (node != null) {
length++;
node = node.next;
}
return length;
}
}
