Linked List Deletion - Amazon Top Interview Questions


Problem Statement :


Given a singly linked list node, and an integer target, return the same linked list with all nodes whose value is target removed.

Constraints

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


Example 1

Input

node = [0, 1, 1, 2]
target = 1


Output

[0, 2]



Solution :



title-img




                        Solution in C++ :

** Recursive

LLNode* solve(LLNode* node, int target) {
    if(not node) return node;
    node->next = solve(node->next, target);
    return node->val == target ? node->next : node;
}


** Iterative


LLNode* solve(LLNode* head, int target) {
    auto* node = head;
    while (node) {
        while (node->next && node->next->val == target) {
            node->next = node->next->next;
        }
        node = node->next;
    }
    return (head->val == target ? head->next : head);
 }
                    


                        Solution in Java :

import java.util.*;

/**
 * class LLNode {
 *   int val;
 *   LLNode next;
 * }
 */
class Solution {
    public LLNode solve(LLNode node, int target) {
        LLNode prev = node;
        LLNode cur = node.next;
        while (cur != null) {
            LLNode nxt = cur.next;
            if (cur.val == target)
                prev.next = nxt;
            else {
                prev = cur;
            }
            cur = nxt;
        }
        if (node.val == target)
            return node.next;
        return node;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, node, target):
        while node != None and node.val == target:
            node = node.next
        ret = node
        while node != None:
            while node.next != None and node.next.val == target:
                node.next = node.next.next
            node = node.next
        return ret
                    


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