Write a program to find the node at which the intersection of two singly linked lists begins.
Example:
The following two linked lists:
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| 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
(Intersection-of-Two-Linked-Lists.java) download
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| /**
* 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 a = headA;
ListNode b = headB;
while(a!=b){
if(a==null)
a = headB;
else
a = a.next;
if(b==null)
b = headA;
else
b = b.next;
}
return a;
}
}
|
(Intersection-of-Two-Linked-Lists.py) download
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| # Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
# find lenA and lenB
lenA = self.getsize(headA)
lenB = self.getsize(headB)
# find intersection
if lenA<lenB:
return self.getIntersectionNode(headB,headA)
for i in range(lenA-lenB):
headA = headA.next
while headA and headB:
if headA.val == headB.val:
return headA
headA = headA.next
headB = headB.next
return None
def getsize(self, node):
len = 0
while node:
len += 1
node = node.next
return len
|