Intersection of Two Linked Lists

lc160

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.

这个解法有意思，先a1走到头，然后将a1的尾巴和b的头帘起来，然后找a的cycle 就行了。

public class Solution {
    /**
     * @param headA: the first list
     * @param headB: the second list
     * @return: a ListNode 
     */
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }

        // get the tail of list A.
        ListNode node = headA;
        while (node.next != null) {
            node = node.next;
        }
        node.next = headB;
        ListNode result = listCycleII(headA);
        node.next = null;
        return result;
    }

    private ListNode listCycleII(ListNode head) {
        ListNode slow = head, fast = head.next;

        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return null;
            }

            slow = slow.next;
            fast = fast.next.next;
        }

        slow = head;
        fast = fast.next;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }

        return slow;
    }
}

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    //boundary check
    if(headA == null || headB == null) return null;

    ListNode a = headA;
    ListNode b = headB;

    //if a & b have different len, then we will stop the loop after second iteration
    while( a != b){
        //for the end of first iteration, we just reset the pointer to the head of another linkedlist
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }

    return a;
}

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). Space complexity : 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). Space complexity : O(m)O(m) or 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). Space complexity : O(1)O(1).