Linked list related interview questions

1) Enter the chain header node. The odd length returns the midpoint, and the even length returns the upper midpoint
2) Enter the chain header node. The odd length returns the midpoint, and the even length returns the lower midpoint
3) Enter the chain header node. The odd length returns the previous midpoint, and the even length returns the previous midpoint
4) Enter the chain header node. The odd length returns the one before the midpoint, and the even length returns the one before the next midpoint

Tips:

Find the upper midpoint. Only two nodes after the current node continue to jump

Find the lower midpoint. If there is a node after the current node, continue to jump

/** * Problem Description: speed pointer * <p> * 1)Enter the chain header node. The odd length returns the midpoint, and the even length returns the upper midpoint * 2)Enter the chain header node. The odd length returns the midpoint, and the even length returns the lower midpoint * 3)Enter the chain header node. The odd length returns the previous midpoint, and the even length returns the previous midpoint * 4)Enter the chain header node. The odd length returns the one before the midpoint, and the even length returns the one before the next midpoint */ public class LinkedListMid { public static class Node { public int value; public Node next; public Node(int value) { this.value = value; } } //1. Midpoint or upper midpoint public static Node midOrUpMidNode(Node head) { if (head == null || head.next == null || head.next.next == null) { return head; } Node slow = head.next;//Slow pointer Node fast = head.next.next;//Quick pointer while (fast.next != null && fast.next.next != null) { fast = fast.next.next; slow = slow.next; } return slow; } //2. Midpoint or lower midpoint public static Node midOrDownMidNode(Node head) { if (head == null || head.next == null) { return head; } Node slow = head.next; Node fast = head.next.next; while (fast != null && fast.next != null) { fast = fast.next.next; slow = slow.next; } return slow; } //3. Return to the one before the midpoint or the one before the upper midpoint (slow one cloth less) public static Node midOrUpMidPreNode(Node head) { if (head == null || head.next == null || head.next.next == null) { return null; } Node slow = head; Node fast = head.next.next; while (fast.next != null && fast.next.next != null) { fast = fast.next.next; slow = slow.next; } return slow; } //4. Return to the one before the midpoint or the one before the next midpoint public static Node midOrDownMidPreNode(Node head) { if (head == null || head.next == null) { return null; } Node slow = head; Node fast = head.next.next; while (fast != null && fast.next != null) { fast = fast.next.next; slow = slow.next; } return slow; } }

Question Description: whether it is palindrome structure. Given the head node of a single linked list, please judge whether the linked list is palindrome structure.

For example: 12321 √ 12312 ×

Prompt: 1 - > 2 - > 3 < - 2 < - 1
↓
null

package com.lzf2.class08; import java.util.Stack; //Whether it is palindrome structure. Given the head node of a single linked list, please judge whether the linked list is palindrome structure. public class IsPalindromeList { public static class Node{ public int value; public Node next; public Node(int value) { this.value = value; } } // need n extra space public static boolean isPalindrome1(Node head) { Stack<Node> stack = new Stack<Node>(); Node cur = head; while (cur != null) { stack.push(cur); cur = cur.next; } while (head != null) { if (head.value != stack.pop().value) { return false; } head = head.next; } return true; } // need n/2 extra space public static boolean isPalindrome2(Node head) { if (head == null || head.next == null) { return true; } Node right = head.next; Node cur = head; while (cur.next != null && cur.next.next != null) { right = right.next; cur = cur.next.next; } Stack<Node> stack = new Stack<Node>(); while (right != null) { stack.push(right); right = right.next; } while (!stack.isEmpty()) { if (head.value != stack.pop().value) { return false; } head = head.next; } return true; } // need O(1) extra space public static boolean isPalindrome3(Node head) { if (head == null || head.next == null){ return true; } //1. The speed pointer finds the middle position of the linked list Node slow = head;//midpoint Node fast = head; while (fast.next != null && fast.next.next != null){ fast = fast.next.next; slow = slow.next; } //2. Adjust the linked list structure Node curr = slow.next; slow.next = null; while (curr != null){ Node next = curr.next; curr.next = slow; slow = curr; curr = next; } //curr = null,slow in the last //3. For comparison, pay attention to slow. You need to readjust the linked list later Node R = slow; Node L = head; boolean res = true; while (L != null && R != null){ if(L.value != R.value){ res = false; break;//Don't end first. You need to adjust the linked list back } L = L.next; R = R.next; } //Tail slow curr = slow; Node pre = null; while (curr != null){ Node next = curr.next; curr.next = pre; pre = curr; curr = next; } return res; } public static void printLinkedList(Node node) { System.out.print("Linked List: "); while (node != null) { System.out.print(node.value + " "); node = node.next; } System.out.println(); } public static void main(String[] args) { Node head = null; printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(2); head.next.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(2); head.next.next.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); } }

Problem Description: divide the one-way linked list into the form of small on the left, equal in the middle and large on the right according to a certain value

Tip: the values are divided into smaller than the table, equal to the table, and larger than the table. Record the header and footer respectively.
Finally, connect the three linked lists

//The one-way linked list is divided into the form of small on the left, equal in the middle and large on the right according to a certain value public class SmallerEqualBigger { //Node node public static class Node { public int value; public Node next; public Node(int value) { this.value = value; } } public static Node listPartition(Node head, int pivot) { Node sH = null;//small head Node sT = null;//small tail Node eH = null;//equal head Node eT = null;//equal tail Node mH = null;//big head Node mT = null;//big tail Node cur = head; while (cur != null) { Node next = cur.next; cur.next = null; if (cur.value > pivot) { if(mH == null){ mH = cur; mT = cur; }else { mT.next = cur; mT = cur; } } else if (cur.value < pivot) { if (sH == null){ sH = cur; sT = cur; }else { sT.next = cur; sT = cur; } } else { if (eH == null){ eH = cur; eT = cur; }else { eT.next = cur; eT = cur; } } cur = next; } //Put the three areas together if (sT != null){ sT.next = eH; eT = eH == null ? sT : eT; } if (eT != null){ eT.next = mH; } return sH != null ? sH : (eH != null ?eH : mH); } public static void printLinkedList(Node node) { System.out.print("Linked List: "); while (node != null) { System.out.print(node.value + " "); node = node.next; } System.out.println(); } public static void main(String[] args) { Node head1 = new Node(7); head1.next = new Node(9); head1.next.next = new Node(1); head1.next.next.next = new Node(8); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(2); head1.next.next.next.next.next.next = new Node(5); printLinkedList(head1); // head1 = listPartition1(head1, 4); head1 = listPartition(head1, 5); printLinkedList(head1); } }

Problem Description:

A special single linked list node class is described below
class Node {
int value;
Node next;
Node rand;
Node(int val) { value = val; }
}
rand pointer is a new pointer in the single linked list node structure. rand may point to any node in the linked list or null.
Given the head Node of an acyclic single linked list composed of Node types, please implement a function to copy the linked list and return the head Node of the copied new linked list.
[requirements]
Time complexity O(N), additional space complexity O(1)

Tip: copy each node and put it behind the original node. Then connect the rand pointer, and finally remove the pointer of the original node.

import java.util.HashMap; //Copy a special linked list (with random pointer) public class CopyListWithRandom { public static class Node { int value; Node next; Node rand; public Node(int value) { this.value = value; } } public static Node copyListWithRand2(Node head) { if (head == null){ return null; } //1.copy each new node is placed behind the old node Node cur = head; while (cur != null){ Node next = cur.next; cur.next = new Node(cur.value); cur.next.next = next; cur = next; } //2. Traverse two nodes at a time, and connect the random pointer of the new node cur = head; while (cur != null){ cur.next.rand = cur.rand != null ? cur.rand.next : null; cur = cur.next.next; } //3. Separate the new node from the old node Node res = head.next; cur = head; while (cur != null){ Node next = cur.next.next; Node copy = cur.next; cur.next = next; copy.next = next != null ? next.next : null; cur = next; } return res; } public static Node copyListWithRand1(Node head) { // key old node // value new node HashMap<Node, Node> map = new HashMap<Node, Node>(); Node cur = head; while (cur != null) { map.put(cur, new Node(cur.value)); cur = cur.next; } cur = head; while (cur != null) { // cur old // map.get(cur) // New. Next - > cur.next clone node found map.get(cur).next = map.get(cur.next); map.get(cur).rand = map.get(cur.rand); cur = cur.next; } return map.get(head); } public static void printRandLinkedList(Node head) { Node cur = head; System.out.print("order: "); while (cur != null) { System.out.print(cur.value + " "); cur = cur.next; } System.out.println(); cur = head; System.out.print("rand: "); while (cur != null) { System.out.print(cur.rand == null ? "- " : cur.rand.value + " "); cur = cur.next; } System.out.println(); } public static void main(String[] args) { Node head = null; Node res1 = null; Node res2 = null; printRandLinkedList(head); res1 = copyListWithRand1(head); printRandLinkedList(res1); res2 = copyListWithRand2(head); printRandLinkedList(res2); printRandLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); head.next.next.next.next.next = new Node(6); head.rand = head.next.next.next.next.next; // 1 -> 6 head.next.rand = head.next.next.next.next.next; // 2 -> 6 head.next.next.rand = head.next.next.next.next; // 3 -> 5 head.next.next.next.rand = head.next.next; // 4 -> 3 head.next.next.next.next.rand = null; // 5 -> null head.next.next.next.next.next.rand = head.next.next.next; // 6 -> 4 System.out.println("Original linked list:"); printRandLinkedList(head); System.out.println("========================="); res1 = copyListWithRand1(head); System.out.println("Copy linked list of method 1:"); printRandLinkedList(res1); System.out.println("========================="); res2 = copyListWithRand2(head); System.out.println("Copy linked list of method 2:"); printRandLinkedList(res2); System.out.println("========================="); System.out.println("Original linked list after method 2 copy:"); printRandLinkedList(head); System.out.println("========================="); } }

Given two single linked lists that may or may not have rings, the head nodes head1 and head2. Please implement a function. If two linked lists intersect, please return the first node of the intersection. If it does not intersect, null is returned
[requirements]
If the sum of the lengths of the two linked lists is N, the time complexity should reach O(N) and the additional space complexity should reach O(1).

Tips:

Two linked lists: 1) both have no rings, 2) both have rings, 3) one has a ring and the other has no ring
First judge whether the linked list has a ring
Provide processing methods for cases 1 and 2 to find the intersection node, and there must be no intersection point in case 3

/** * Given two single linked lists that may or may not have rings, the head nodes head1 and head2. Please implement a function. If two linked lists intersect, please return the first node of the intersection. If it does not intersect, null is returned * [Requirements] * If the sum of the lengths of the two linked lists is N, the time complexity should reach O(N) and the additional space complexity should reach O(1). */ public class FindFirstIntersectNode { public static class Node{ public int value; public Node next; public Node(int value) { this.value = value; } } //Returns the intersection node of two linked lists, and returns null if there is no intersection public static Node getIntersectNode(Node head1,Node head2){ if (head1 == null || head2 == null){ return null; } Node loop1 = getLoopNode(head1); Node loop2 = getLoopNode(head2); if(loop1 == null && loop2 == null){ return noLoop(head1,head2); } if(loop1 != null && loop2 != null){ return bothLoop(head1,loop1,head2,loop2); } return null; } //Find the first ring entry node of the two linked lists. If there is no ring, null will be returned. public static Node getLoopNode(Node head){ if(head == null || head.next == null || head.next.next == null){ return null; } //Speed pointer Node fast = head.next.next; Node slow = head.next; while (fast != slow){ if(fast.next == null || fast.next.next == null){ return null; } fast = fast.next.next; slow = slow.next; } //fast and slow met fast = head; while (fast != slow){ fast = fast.next; slow = slow.next; } return slow; } //Handle the case where both are acyclic public static Node noLoop(Node head1,Node head2){ if (head1 == null || head2 == null){ return null; } Node cur1 = head1; Node cur2 = head2; int n = 0; while (cur1.next != null){ n++; cur1 = cur1.next; } while (cur2.next != null){ n--; cur2 = cur2.next; } //The tail nodes of the two linked lists are different and do not intersect if(cur1 != cur2){ return null; } //Find out the long node and take the difference step (step n) cur1 = n > 0 ? head1 : head2;//Long cur2 = cur1 == head1 ? head2 : head1;//Short n = Math.abs(n); while (n != 0){ cur1 = cur1.next; n--; } //Two linked lists go together while (cur1 != cur2){ cur1 = cur1.next; cur2 = cur2.next; } return cur1; } //Handle the case where both have rings public static Node bothLoop(Node head1,Node loop1,Node head2,Node loop2){ Node cur1 = null; Node cur2 = null; if (loop1 == loop2){//Same entry point cur1 = head1; cur2 = head2; int n = 0; while (cur1 != loop1){ n++; cur1 = cur1.next; } while (cur2 != loop2){ n--; cur2 = cur2.next; } cur1 = n > 0 ? head1 : head2;//long cur2 = cur1 == head1 ? head2 : head1; n = Math.abs(n); //Take the difference step while (n != 0){ cur1 = cur1.next; n--; } //Two go together while (cur1 != cur2){ cur1 = cur1.next; cur2 = cur2.next; } return cur1; }else { Node cur = loop1.next; while (cur != loop1){//Walk around if(cur == loop2){ return loop1; } cur = cur.next; } return null;//Two rings do not intersect } } public static void main(String[] args) { // 1->2->3->4->5->6->7->null Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(6); head1.next.next.next.next.next.next = new Node(7); // 0->9->8->6->7->null Node head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next.next.next.next.next; // 8->6 System.out.println(getIntersectNode(head1, head2).value); // 1->2->3->4->5->6->7->4... head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(6); head1.next.next.next.next.next.next = new Node(7); head1.next.next.next.next.next.next = head1.next.next.next; // 7->4 // 0->9->8->2... head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next; // 8->2 System.out.println(getIntersectNode(head1, head2).value); // 0->9->8->6->4->5->6.. head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next.next.next.next.next; // 8->6 System.out.println(getIntersectNode(head1, head2).value); } }

Problem Description: delete a node without a header node

It can be implemented, but it is not really deleting nodes, but only changing values, which will cause problems in some scenarios.
And the last node cannot be deleted.

public static void delete(Node node){ if(node == null){ return; } Node next = node.next; if(next == null){//The current node is the last node and cannot be deleted return; } node.value = next.value; node.next = next.next; }

14 October 2021, 18:02 | Views: 2726