Back to Front Linked List - Google Top Interview Questions


Problem Statement :


Given a singly linked list node, reorder it such that we take: the last node, and then the first node, and then the second last node, and then the second node, etc.

Can you do it in \mathcal{O}(1)O(1) space?

Constraints

0 ≤ n ≤ 100,000 where n is the number of nodes in node

Example 1

Input



node = [0, 1, 2, 3]

Output

[3, 0, 2, 1]


Solution :



title-img 




                        Solution in C++ :

LLNode* solve(LLNode* node) {
    LLNode* slow = node;
    LLNode* fast = node;
    while (fast) {
        fast = fast->next;
        if (fast) {
            fast = fast->next;
            slow = slow->next;
        }
    }
    LLNode* pre = NULL;
    LLNode* cur = slow;
    LLNode* nex;
    while (cur) {
        nex = cur->next;
        cur->next = pre;
        pre = cur;
        cur = nex;
    }
    LLNode* temp = pre;
    while (temp && node) {
        LLNode* nex1 = temp->next;
        LLNode* nex2 = node->next;
        temp->next = node;
        node->next = nex1;
        temp = nex1;
        node = nex2;
    }
    return pre;
}
                    


                        Solution in Java :

import java.util.*;

/**
 * class LLNode {
 *   int val;
 *   LLNode next;
 * }
 */
class Solution {
    public LLNode solve(LLNode node) {
        LLNode fast = node, slow = node;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        LLNode left = node, right = reverse(slow);
        LLNode prev = new LLNode();
        LLNode dummy = prev;
        while (left != null && right != null) {
            prev.next = right;
            right = right.next;
            prev.next.next = left;
            left = left.next;
            prev = prev.next.next;
        }
        prev.next = null;
        return dummy.next;
    }

    private LLNode reverse(LLNode node) {
        LLNode prev = null, curr = node, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def split(self, head):
        tail1 = head
        tail2 = head

        while tail1 and tail2 and tail2.next and tail2.next.next:
            tail1 = tail1.next
            tail2 = tail2.next.next

        # now actually break the link!
        head2 = tail1.next
        tail1.next = None

        return head, head2

    def reverse(self, head):
        if not head or not head.next:
            return head
        prev, cur = None, head
        while cur:
            nxt = cur.next
            cur.next = prev
            prev = cur
            cur = nxt

        return prev

    def interleave(self, head, head2):
        resHead = res = LLNode(0)
        while head or head2:
            if head2:
                res.next = head2
                res = res.next
                head2 = head2.next

            if head:
                res.next = head
                res = res.next
                head = head.next

        return resHead.next

    def solve(self, head):
        # break into two halves
        head, head2 = self.split(head)

        # now reverse the second list ;)
        head2 = self.reverse(head2)

        # just alternately pick from each
        return self.interleave(head, head2)


View More Similar Problems

Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

View Solution →

Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

View Solution →

Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →

Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →

The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →

Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →