19. Remove Nth Node From End of List
Given a linked list, remove the n-th node from the end of list and return its head.
Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
題意：給出一個鏈表，數n，將這個鏈表的倒數第n個數刪除，然後在把這個鏈表輸出
思路:
使用兩個快慢指針,一個slow,一個fast
首先fast前進n個節點
然後slow從頭(head)和fast一起每次前進一個節點,當fast鏈表尾部的時候,
slow指向的就是倒數第n個節點
（看別人的博客）
JAVA
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode fast=head;
for(int i=0;i<n;i++)
{
fast=fast.next;
}
if(fast==null)
{
return head.next;
}
ListNode slow=head;
while(fast.next!=null)
{
slow=slow.next;
fast=fast.next;
}
slow.next=slow.next.next;
return head;
}
}
234. Palindrome Linked List
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2
Output: false
Example 2:
Input: 1->2->2->1
Output: true
思路：快慢指針找到中間的位置，然後將後半部分的鏈表反轉，然後跟前半部分的鏈表逐個位置比對
JAVA
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isPalindrome(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null){
fast = fast.next.next;
slow =slow.next;
}
if(fast != null)
slow =slow.next;
slow = reverseList(slow);
while(slow != null){
if(slow.val != head.val)
return false;
slow = slow.next;
head = head.next;
}
return true;
}
public ListNode reverseList(ListNode head){
ListNode pre = null, next = null;
while(head != null){
next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
JAVA2
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isPalindrome(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null){
fast = fast.next.next;
slow =slow.next;
}
if(fast != null)
slow =slow.next;
slow = reverseList(slow);
while(slow != null){
if(slow.val != head.val)
return false;
slow = slow.next;
head = head.next;
}
return true;
}
public ListNode reverseList(ListNode head) {
if(head==null)
{
return head;
}
ListNode p=head;
ListNode q=head;
while(head.next!=null)
{
p=head.next;
head.next=head.next.next;
p.next=q;
q=p;
}
return q;
}
}
21. Merge Two Sorted Lists
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
JAVA
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1==null) return l2;
if(l2==null) return l1;
if(l1.val < l2.val)
{
l1.next=mergeTwoLists(l1.next,l2);
return l1;
}else
{
l2.next=mergeTwoLists(l2.next,l1);
return l2;
}
}
}
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1==NULL) return l2;
if(l2==NULL) return l1;
if(l1->val < l2->val)
{
l1->next=mergeTwoLists(l1->next,l2);
return l1;
}else
{
l2->next=mergeTwoLists(l2->next,l1);
return l2;
}
}
};
141. Linked List Cycle
Given a linked list, determine if it has a cycle in it.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
題目要求是判斷鏈表是否有環
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
ListNode *p=head;
ListNode *q=head;
while(p!=NULL && p->next!=NULL )
{
q=q->next;
p=p->next->next;
if(q==p) return true;
}
return false;
}
};
JAVA
/**
* 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) {
ListNode p=head;
ListNode q=head;
while(p!=null && p.next!=null )
{
q=q.next;
p=p.next.next;
if(q==p) return true;
}
return false;
}
}
206. Reverse Linked List
Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head==NULL)
{
return head;
}
ListNode *p=head;
ListNode *q=head;
while(head->next!=NULL)
{
p=head->next;
head->next=head->next->next;
p->next=q;
q=p;
}
return q;
}
};
JAVA
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if(head==null)
{
return head;
}
ListNode p=head;
ListNode q=head;
while(head.next!=null)
{
p=head.next;
head.next=head.next.next;
p.next=q;
q=p;
}
return q;
}
}
876. Middle of the Linked List
Given a non-empty, singly linked list with head node head, return a middle node of linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.
JAVA
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode middleNode(ListNode head) {
if(head==null || head.next==null)
return head;
ListNode slow=head;
ListNode fast=head;
fast=fast.next.next;
while(fast!=null && fast.next!=null){
slow=slow.next;
fast=fast.next.next;
}
return slow.next;
}
}
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* middleNode(ListNode* head) {
if(head==NULL || head->next==NULL)
return head;
ListNode* slow=head;
ListNode* fast=head;
fast=fast->next->next;
while(fast!=NULL && fast->next!=NULL){
slow=slow->next;
fast=fast->next->next;
}
return slow->next;
}
};