160. Intersection of Two Linked Lists (Easy)

https://leetcode.com/problems/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.

Solutions


class Solution {
    // This problem is easy to solve. The key point is to figure out the length of both list.

    // It's clear that once both list move forward simultaneous, only if they have same length
    // so that they can meet at the intersect node. The trick is that we can let the longer one
    // move several steps in advance before both set off. The advanced steps should be the
    // difference of the longer deducting the shorter. Now they are at the same start line and
    // the distance to intersect node is the same.

    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }

        int lenA = length(headA);
        int lenB = length(headB);

        ListNode ptrA = headA;
        ListNode ptrB = headB;

        while (lenA > lenB) {
            ptrA = ptrA.next;
            lenA--;
        }

        while (lenB > lenA) {
            ptrB = ptrB.next;
            lenB--;
        }

        while (ptrA != null && ptrB != null) {
            if (ptrA == ptrB) {
                return ptrB;
            }

            ptrA = ptrA.next;
            ptrB = ptrB.next;
        }

        return null;
    }

    private int length(ListNode head) {
        int l = 0;

        while (head != null) {
            l++;
            head = head.next;
        }

        return l;
    }
}

Incorrect Solutions

References

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