203. Remove Linked List Elements
Solved
Easy
Topics
Companies文章來源：http://www.zghlxwxcb.cn/news/detail-848916.html
Given the?head?of a linked list and an integer?val, remove all the nodes of the linked list that has?Node.val == val, and return?the new head.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode dummyHead = new ListNode(-1,head);
ListNode cur = dummyHead;
while(cur != null && cur.next != null){
if(cur.next.val == val){
cur.next = cur.next.next;
}else{
cur = cur.next;
}
}
return dummyHead.next;
}
}206. Reverse Linked List
Solved
Easy
Topics
Companies
Given the?head?of a singly linked list, reverse the list, and return?the reversed list.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode newHead = null;
ListNode o1 = head;
while(o1 != null){
ListNode o2 = o1.next;
o1.next = newHead;
newHead = o1;
o1 = o2;
}
return newHead;
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode last = reverseList(head.next);
head.next.next = head;
head.next = null;
return last;
}
}24. Swap Nodes in Pairs
Solved
Medium
Topics
Companies
Given a?linked list, swap every two adjacent nodes and return its head. You must solve the problem without?modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
ListNode dummyHead = new ListNode(-1 , head);
ListNode cur = dummyHead;
ListNode n1 = head;
while(n1 != null && n1.next != null){
ListNode n2 = n1.next.next;
cur.next = n1.next;
n1.next.next = n1;
n1.next = n2;
cur = n1;
n1 = n2;
}
return dummyHead.next;
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode swapPairs(ListNode head) {
if(head == null||head.next==null){
return head;
}
ListNode newHead = head.next;
head.next = swapPairs(newHead.next);
newHead.next = head;
return newHead;
}
}
19. Remove Nth Node From End of List
Solved
Medium
Topics
Companies
Hint
Given the?head?of a linked list, remove the?nth?node from the end of the list and return its head.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummyHead = new ListNode(-1, head);
ListNode n1 = dummyHead;
ListNode n2 = dummyHead;
for(int i = 0; i < n + 1; i++){
n2 = n2.next;
}
while(n2 != null){
n2 = n2.next;
n1 = n1.next;
}
n1.next = n1.next.next;
return dummyHead.next;
}
}141. Linked List Cycle
Easy
Given?head, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the?next?pointer. Internally,?pos?is used to denote the index of the node that?tail's?next?pointer is connected to.?Note that?pos?is not passed as a parameter.
Return?true?if there is a cycle in the linked list. Otherwise, return?false.
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
if (head == null || head.next == null) {
return false;
}
ListNode slow = head;
ListNode fast = head.next;
while(fast!=null && fast.next!=null){
if(slow == fast) return true;
slow=slow.next;
fast=fast.next.next;
}
return false;
}
}142. Linked List Cycle II
Solved
Medium
Topics
Companies
Given the?head?of a linked list, return?the node where the cycle begins. If there is no cycle, return?null.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the?next?pointer. Internally,?pos?is used to denote the index of the node that tail's?next?pointer is connected to (0-indexed). It is?-1?if there is no cycle.?Note that?pos?is not passed as a parameter.
Do not modify?the linked list.
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head==null){
return null;
}
ListNode fast = head;
ListNode slow =head;
while(fast!=null && fast.next!=null){
fast = fast.next.next;
slow = slow.next;
if(slow == fast){
slow = head;
while(slow!=fast){
slow = slow.next;
fast = fast.next;
}
return slow;
}
}
return null;
}
}21. Merge Two Sorted Lists
Solved
Easy
Topics
Companies
You are given the heads of two sorted linked lists?list1?and?list2.
Merge the two lists into one?sorted?list. The list should be made by splicing together the nodes of the first two lists.
Return?the head of the merged linked list.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode dummyHead = new ListNode(-1,null);
ListNode n = dummyHead;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
n.next = list1;
list1 = list1.next;
} else {
n.next = list2;
list2 = list2.next;
}
n = n.next;
}
if (list1 == null) {
n.next = list2;
}
if (list2 == null) {
n.next = list1;
}
return dummyHead.next;
}
}82. Remove Duplicates from Sorted List II
Solved
Medium
Topics
Companies
Given the?head?of a sorted linked list,?delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return?the linked list?sorted?as well.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode dummyHead = new ListNode(-1 , head);
ListNode slow = dummyHead;
ListNode fast = head;
while(fast != null && fast.next != null){
if(fast.val == fast.next.val){
int val = fast.val;
while(fast != null && fast.val == val){
fast = fast.next;
slow.next = fast;
}
}else{
fast = fast.next;
slow = slow.next;
}
}
return dummyHead.next;
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode dummyHead = new ListNode(-1, head);
ListNode cur = dummyHead;
while (cur.next != null && cur.next.next != null) {
if (cur.next.val == cur.next.next.val) {
ListNode p = cur.next;
while (cur.next != null && cur.next.val == p.val) {
cur.next = cur.next.next;
}
} else {
cur = cur.next;
}
}
return dummyHead.next;
}
}83. Remove Duplicates from Sorted List
Solved
Easy
Topics
Companies
Given the?head?of a sorted linked list,?delete all duplicates such that each element appears only once. Return?the linked list?sorted?as well.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode cur = head;
while(cur != null && cur.next != null){
if(cur.next.val == cur.val){
cur.next = cur.next.next;
}else{
cur = cur.next;
}
}
return head;
}
}160. Intersection of Two Linked Lists
Solved
Easy
Topics
Companies
Given the heads of two singly linked-lists?headA?and?headB, return?the node at which the two lists intersect. If the two linked lists have no intersection at all, return?null.文章來源地址http://www.zghlxwxcb.cn/news/detail-848916.html
/**
* 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) {
ListNode curA = headA;
ListNode curB = headB;
while(curA != curB){
if(curA != null){
curA = curA.next;
}else{
curA = headB;
}
if(curB != null){
curB = curB.next;
}else{
curB = headA;
}
}
return curA;
}
}到了這里，關(guān)于Linked List Mock的文章就介紹完了。如果您還想了解更多內(nèi)容，請在右上角搜索TOY模板網(wǎng)以前的文章或繼續(xù)瀏覽下面的相關(guān)文章，希望大家以后多多支持TOY模板網(wǎng)！
