Question

Formatted question description: https://leetcode.ca/all/160.html

 160	Intersection of Two Linked Lists

 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.

 @tag-linkedlist

Algorithm

If the two chains are the same length, then the corresponding ones can be found one by one, so you only need to shorten the long list.

The specific algorithm is: traverse the two linked lists respectively to obtain the corresponding lengths. Then find the difference in length, move the longer linked list backward by the number of differences, and then compare them one by one.

Code

Java

  • 
    public class Intersection_of_Two_Linked_Lists {
        /**
         * Definition for singly-linked list.
         * public class ListNode {
         *     int val;
         *     ListNode next;
         *     ListNode(int x) {
         *         val = x;
         *         next = 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) {
    
                int lenA = getLength(headA);
                int lenB = getLength(headB);
    
                // move headA and headB to the same start point
                while (lenA > lenB) {
                    headA = headA.next;
                    lenA--;
                }
                while (lenA < lenB) {
                    headB = headB.next;
                    lenB--;
                }
    
                // find the intersection until end
                while (headA != headB) { // same reference
                    headA = headA.next;
                    headB = headB.next;
                }
                return headA;
            }
    
            private int getLength(ListNode node) {
                int length = 0;
                while (node != null) {
                    node = node.next;
                    length++;
                }
                return length;
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/intersection-of-two-linked-lists/
    // Time: O(A + B)
    // Space: O(1)
    class Solution {
        int getLength(ListNode *h) {
            int len = 0;
            for (; h; h = h->next) ++len;
            return len;
        }
    public:
        ListNode *getIntersectionNode(ListNode *a, ListNode *b) {
            int la = getLength(a), lb = getLength(b);
            if (la < lb) swap(a, b), swap(la, lb);
            for (int i = la - lb; i > 0; --i) a = a->next;
            while (a && b && a != b) a = a->next, b = b->next;
            return a;
        }
    };
    
  • class Solution:
      def getIntersectionNode(self, headA, headB):
        if headA is None or headB is None:
          return None
        pa = headA
        pb = headB
        while pa is not pb:
          pa = headB if pa is None else pa.next
          pb = headA if pb is None else pb.next
        return pa
    
    

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