160. Intersection of Two Linked Lists
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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.
官方解答:
Solution
Approach #1 (Brute Force) [Time Limit Exceeded]
For each node ai in list A, traverse the entire list B and check if any node in list B coincides with ai.
Complexity Analysis
- Time complexity : O(mn)O(mn)O(mn).
- Space complexity : O(1)O(1)O(1).
Approach #2 (Hash Table) [Accepted]
Traverse list A and store the address / reference to each node in a hash set. Then check every node bi in list B: if bi appears in the hash set, then bi is the intersection node.
Complexity Analysis
- Time complexity : O(m+n)O(m+n)O(m+n).
- Space complexity : O(m)O(m)O(m) or O(n)O(n)O(n).
Approach #3 (Two Pointers) [Accepted]
- Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
- When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
- If at any point pA meets pB, then pA/pB is the intersection node.
- To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
- If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
Complexity Analysis
- Time complexity : O(m+n)O(m+n)O(m+n).
- Space complexity : O(1)O(1)O(1).
Analysis written by @stellari.
源码:
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode first,second;
first=headA;
second=headB;
while(first!=second)
{
if(first!=null)
{
first=first.next;
}
else
{
first=headB;
}
if(second!=null)
{
second=second.next;
}
else
{
second=headA;
}
}
return first;
}
}
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