链表数据结构：
public class LinkedList{
int val;
LinkedList next;
public LinkedList(int val){this.val = val;}
}
1\翻转链表 Leetcode 206
关键为对下一个next的备份，对上一个节点的记录（temp）
/**
* 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;
if(head.next == null) return head;
ListNode next; //备份
ListNode temp = null;
while(head!=null)
{
next=head.next;
head.next = temp;
temp=head;
head=next;
}
return temp;
}
}
2\翻转链表2 Leetcode 92
反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if(head.next==null) return head;
ListNode temphead = head;
ListNode before = null;
for(int i=0;i<m-1;i++)
{
before = temphead;
temphead = temphead.next;
}
ListNode begin = temphead;
ListNode temp = null;
ListNode tempbf = null;
for(int i=0;i<n-m+1;i++)
{
temp = temphead.next;
temphead.next=tempbf;
tempbf=temphead;
temphead=temp;
}
begin.next=temphead;
if(before==null) return tempbf; //注意是不是从第一个开始翻转
else before.next = tempbf;
return head;
}
}
3\求两个链表的交点 Leetcode 160 https://leetcode-cn.com/problems/intersection-of-two-linked-lists/
方法一 暴力法：
/**
* 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 a = headA;
ListNode b = headB;
while(a!=null)
{
b = headB;
while(b!=null)
{
if(a==b)
{
return b;
}
b = b.next;
}
a = a.next;
}
return null;
}
}
方法二 快慢指针：
/**
* 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 a = headA;
ListNode b = headB;
int lena =0;
int lenb =0;
while(a!=null)
{
lena++;
a=a.next;
}
while(b!=null)
{
lenb++;
b=b.next;
}
a=headA;
b=headB;
if(lena>lenb)
{
for(int i=0;i<lena-lenb;i++)
{
a=a.next;
}
}
else
{
for(int i=0;i<lenb-lena;i++)
{
b=b.next;
}
}
if(a==b) return a;
do{
a = a.next;
b = b.next;
}while(a!=b);
if(a==null) return null;
else return a;
}
}
4\链表求环 Leetcode141
给定一个链表，判断链表中是否有环。快慢指针。
/**
* 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 fast = head;
ListNode slow = head;
if(head==null||head.next==null) return false;
while(fast!=null&&slow!=null)
{
fast = fast.next;
if(fast!=null) fast = fast.next;
else return false;
slow = slow.next;
if(slow==fast) return true;
}
return false;
}
}
5\链表求环2 leetcode142
推导 快指针(2)走过的路程是慢指针(1)的两倍 则在h点相遇时 （F+a)*2+n*(a+b)= F+a+b+a+n*(a+b) 则F=b
所以从head与h出发，能找到环的起始节点
/**
* 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) {
ListNode fast = head;
ListNode slow = head;
ListNode meet = null;
if(head==null||head.next==null) return null;
while(fast!=null&&slow!=null)
{
fast=fast.next;
if(fast!=null) fast = fast.next;
else return null;
slow = slow.next;
if(fast==slow)
{
meet=fast;
break;
}
}
if(meet!=null)
{
while(head!=meet)
{
head=head.next;
meet=meet.next;
}
return meet;
}
return null;
}
}
6\分隔链表 leetcode 64
给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。
你应当保留两个分区中每个节点的初始相对位置。
使用临时节点保存
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode small = null;
ListNode smptr = null;
ListNode large = null;
ListNode lgptr = null;
if(head==null) return null;
while(head!=null)
{
if(head.val<x)
{
if(small==null)
{
small = head;
smptr = head;
}
else
{
smptr.next = head;
smptr = head;
}
}
else
{
if(large==null)
{
large = head;
lgptr = head;
}
else{
lgptr.next = head;
lgptr = head;
}
}
head = head.next;
}
if(small==null) return large;
smptr.next = large;
if(large!=null) lgptr.next = null;
return small;
}
}
7\链表深拷贝 太难 leetcode 138
8\链表合并 leetcode 21
将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
/**
* 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) {
//应用临时节点
ListNode newhead = null;
ListNode ptr = null;
if(l1==null) return l2;
if(l2==null) return l1;
while(l1!=null&&l2!=null)
{
if(l1.val<l2.val)
{
if(newhead==null)
{
newhead = l1;
ptr = newhead;
}
else
{
ptr.next = l1;
ptr = l1;
}
l1 = l1.next;
}
else{
if(newhead==null)
{
newhead = l2;
ptr = newhead;
}
else
{
ptr.next = l2;
ptr = l2;
}
l2 = l2.next;
}
}
if(l1==null)
{
ptr.next = l2;
}
else ptr.next = l1;
return newhead;
}
}
9\合并链表2 Leetcode23 https://leetcode-cn.com/problems/merge-k-sorted-lists/
合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。 太难。
原文地址：https://www.cnblogs.com/luiyuying/p/12676299.html
时间： 2024-10-26 22:00:40