160. Intersection of Two Linked Lists
Tags: ‘Linked List’
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 Output: Reference of the node with value = 8 Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 Output: Reference of the node with value = 2 Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
Example 3:
Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 Output: null Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values. Explanation: The two lists do not intersect, so return null.
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 1
Find length first, make two pointer at same position, then traverse at the same time.
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) return null;
int l = length(headA);
int m = length(headB);
while (l > m) {
headA = headA.next;
l--;
}
while (m > l) {
headB = headB.next;
m--;
}
// now A and B are at same position
while (headA != null && headB != null) {
if (headA == headB) return headA;
headA = headA.next;
headB = headB.next;
}
return null;
}
private int length(ListNode node) {
if (node == null) return 0;
int n = 0;
while (node != null) {
n = n + 1;
node = node.next;
}
return n;
}