1、删除链表中的某节点。
Delete Node in a Linked List
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Given linked list -- head = [4,5,1,9], which looks like following:
4 -> 5 -> 1 -> 9
Example 1:
Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list
should become 4 -> 1 -> 9 after calling your function.
Example 2:
Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list
should become 4 -> 5 -> 9 after calling your function.
Note:
The linked list will have at least two elements.
All of the nodes' values will be unique.
The given node will not be the tail and it will always be a valid node of the linked list.
Do not return anything from your function./**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;//连续往后递归，把node.val 移出去
}
}2、删除倒数第N个节点。
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.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?/**
* 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 start = new ListNode(0);
start.next = head;
ListNode slow = start, fast = start;
//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++) {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}
}3、链表反转。
Reverse Linked List
Go to Discuss
Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
int i = 0 ;
ListNode newList = new ListNode();
newList.next = head;
while(head.next != null){
i++;
}
head.next = newList;
return newLsit
}
}4、合并两个排序链表，新链表按数值大小排序。
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/**
* 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(l1, l2.next);
return l2;
}
}
}5、判断链表是否回文。
Palindrome Linked List
Go to Discuss
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
Follow up:
Could you do it in O(n) time and O(1) space?//方法一 可先反转排序之后对比，实际反转一半的链表即可，具体操作见方法二
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isPalindrome(ListNode head) {
ListNode pReversedHead = head;
ListNode pNode = head;
ListNode pPrev = null;
int i = 0;
while (pNode != null) {
ListNode pNext = pNode.next;
i++;
if (pNode.next == null) {
pReversedHead = pNode;
}
pNode.next = pPrev;
pPrev = pNode;
pNode = pNext;
}
ListNode newHead = pReversedHead;
ListNode oldHead = head;
int j = 0;
while(j++<=i/2){
if (newHead == oldHead){
newHead = newHead.next;
oldHead = oldHead.next;
}
else return false;
}
return true;
}
}
//方法二
This can be solved by reversing the 2nd half and compare the two halves. Let's start with an example [1, 1, 2, 1].
In the beginning, set two pointers fast and slow starting at the head.
1 -> 1 -> 2 -> 1 -> null
sf
(1) Move: fast pointer goes to the end, and slow goes to the middle.
1 -> 1 -> 2 -> 1 -> null
s f
(2) Reverse: the right half is reversed, and slow pointer becomes the 2nd head.
1 -> 1 null <- 2 <- 1
h s
(3) Compare: run the two pointers head and slow together and compare.
1 -> 1 null <- 2 <- 1
h s
public boolean isPalindrome(ListNode head) {
ListNode fast = head, slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
if (fast != null) { // odd nodes: let right half smaller
slow = slow.next;
}
slow = reverse(slow);
fast = head;
while (slow != null) {
if (fast.val != slow.val) {
return false;
}
fast = fast.next;
slow = slow.next;
}
return true;
}
public ListNode reverse(ListNode head) {
ListNode prev = null;
while (head != null) {
ListNode next = head.next;
head.next = prev;
prev = head;
head = next;
}
return prev;
}6、检查链表中是否有环。
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?//Use two pointers, walker and runner.
//walker moves step by step. runner moves two steps at time.
//if the Linked List has a cycle walker and runner will meet at some
//point.
public boolean hasCycle(ListNode head) {
if(head==null) return false;
ListNode walker = head;
ListNode runner = head;
while(runner.next!=null && runner.next.next!=null) {
walker = walker.next;
runner = runner.next.next;
if(walker==runner) return true;
}
return false;
}