Alibi

My Life is Getting Better

Leetcode[160] Intersection of Two Linked Lists

14 Nov 2015

###Task1 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.

###Python

# 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
        """
        if not headA or not headB:
            return None
        len1 = 0
        len2 = 0
        itr = headA
        while itr:
            itr = itr.next
            len1 += 1
        itr = headB
        while itr:
            itr = itr.next
            len2 += 1
        if len1 > len2:
            diff = len1 - len2
            while diff > 0:
                headA = headA.next
                diff -= 1
        else:
            diff = len2 - len1
            while diff > 0:
                headB = headB.next
                diff -= 1
        while headA and headB:
            if headA == headB:
                return headA
            headA = headA.next
            headB = headB.next
        return None

###Java ####O(n) space

/**
 * 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;
        }
        HashMap<ListNode, Integer> map = new HashMap<ListNode, Integer>();
        int pos = 0;
        while (headA != null) {
            map.put(headA, pos);
            headA = headA.next;
            pos++;
        }
        while (headB != null) {
            if (map.containsKey(headB)) {
                return headB;
            }
            headB = headB.next;
        }
        
        return null;
    }
}

####O(1) space

/**
 * 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 iterA = headA;
        ListNode iterB = headB;

        int lengthA = 0;
        int lengthB = 0;

        while (iterA != null) {
            lengthA++;
            iterA = iterA.next;
        }
        while (iterB != null) {
            lengthB++;
            iterB = iterB.next;
        }

        iterA = headA;
        iterB = headB;

        if (lengthA > lengthB) {
            int diff = lengthA - lengthB;
            for (int i = 0; i < diff; i++) {
                iterA = iterA.next;
            }
        } else if (lengthB > lengthA) {
            int diff = lengthB - lengthA;
            for (int i = 0; i < diff; i++) {
                iterB = iterB.next;
            }
        } 

        while (iterA != null) {
            if (iterA == iterB) {
                return iterA;
            }
            iterA = iterA.next;
            iterB = iterB.next;
        }

        return null;

    }
}

###Points

Related Posts