目录
1.移除链表元素
2.反转链表
3.链表的中间结点
4.链表中倒数第k个结点
5.合并两个有序链表
6.链表分割
7.链表的回文结构
8.相交链表
9.环形链表
10.环形链表II
1.移除链表元素
链接 203. 移除链表元素 – 力扣(LeetCode)
题目要求
分析一下
上代码
class Solution {public ListNode removeElements(ListNode head, int val) {if (head == null) return null;ListNode prev = head;ListNode cur = head.next;while(cur != null) {if(cur.val == val) {prev.next = cur.next;cur = cur.next;}else {prev = cur;cur = cur.next;}}if (head.val == val) {head= head.next;}return head;}}
2.反转链表
链接 206. 反转链表 – 力扣(LeetCode)
题目要求
分析一下
上代码
class Solution {public ListNode reverseList(ListNode head) {if(head == null) return null;//说明链表为空if(head.next == null) return head;//说明只有一个节点ListNode cur = head.next;head.next = null;while(cur != null) {ListNode curNext= cur.next; cur.next = head;head =cur;cur = curNext;}return head;}}
3.链表的中间结点
链接 876. 链表的中间结点 – 力扣(LeetCode)
题目要求
分析一下
上代码
4.链表中倒数第k个结点
链接 链表中倒数第k个结点_牛客题霸_牛客网 (nowcoder.com)
题目要求
分析一下
上代码
5.合并两个有序链表
链接 21. 合并两个有序链表 – 力扣(LeetCode)
题目要求
分析一下
上代码
class Solution {public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode newHead = new ListNode(-1);//傀儡节点 虚拟结点ListNode tmp = newHead;while(list1 != null && list2 != null) {if(list1.val < list2.val) {tmp.next = list1;list1 = list1.next;tmp = tmp.next;} else {tmp.next = list2;list2 = list2.next;tmp = tmp.next;}}if(list1 != null) {tmp.next = list1;}if(list2 != null) {tmp.next = list2;}return newHead.next;}}
6.链表分割
链接 链表分割_牛客题霸_牛客网 (nowcoder.com)
题目要求
分析一下
上代码
import java.util.*;public class Partition {public ListNode partition(ListNode pHead, int x) {ListNode bs = null;ListNode be = null;ListNode as = null;ListNode ae = null;ListNode cur = pHead;while(cur != null) {if(cur.val < x) {//插入到第一个段if(bs == null) {//这是第一次插入元素bs = cur;be = cur;}else {be.next =cur;be = be.next;}} else {//插入到第二个段if(as == null) {as =cur;ae =cur;}else {ae.next =cur;ae = ae.next;}}cur = cur.next;}if(bs == null) {return as;}be.next =as; if(as != null) {ae.next = null;}return bs;}}
7.链表的回文结构
链接 链表的回文结构_牛客题霸_牛客网 (nowcoder.com)
题目要求
分析一下
上代码
public class PalindromeList {public boolean chkPalindrome(ListNode A) {// write code hereif(A == null || A.next == null) {return true;}ListNode fast = A;ListNode slow = A;//找中点while(fast!= null && fast.next != null) {fast = fast.next.next;slow = slow.next;}//翻转ListNode cur = slow.next;while(cur != null) {ListNode curNext = cur.next;cur.next = slow;slow = cur;cur = curNext;}//对比while(A != slow) {if(A.val != slow.val) {return false;}//偶数情况下if(A.next != slow) { return true;}A = A.next;slow = slow.next;}return true;}}
8.相交链表
链接 160. 相交链表 – 力扣(LeetCode)
题目要求
分析一下
上代码
public class Solution {public ListNode getIntersectionNode(ListNode headA, ListNode headB) {//1.分别求两个链表的长度int lenA = 0;int lenB = 0;ListNode pl = headA;//pl代表永远指向长的链表ListNode ps = headB;//ps代表永远指向短的链表while(pl != null) {lenA++;pl = pl.next;}while(ps != null) {lenB++;ps = ps.next;}pl = headA;ps = headB;//2,求两个链表的差值长度 len一定是一个正数int len = lenA -lenB;if(len < 0) {pl = headB;ps = headA;len = lenB - lenA;}//3.让长链表先走差值长度while(len != 0) {pl = pl.next;len--;}//4.再一块走,相遇while(pl != ps && pl != null) {pl = pl.next;ps = ps.next;}if(pl == null) {return null;}return pl;}}
9.环形链表
链接 141. 环形链表 – 力扣(LeetCode)
题目要求
分析一下
首先,明白什么叫链表带环,
由链表中的某一个节点,一直找next指针,一定会再次到达这个节点,这样的链表就说是有环的,并且带环的链表每个节点的next 指针,都不为null。
其次,这道题如果继续考虑使用快慢指针来做,那么快指针fast每次走几步,
(1)如果快指针fast每次都两步,慢指针slow每次走一步,一定可以相遇
(2)如果快指针fast每次都走3步,走4步…走n步,不一定可以相遇
所以只要快指针fast每次走2步,慢指针每次走1步,就一定可以追上,这道题就按这个思路走
上代码
public class Solution {public boolean hasCycle(ListNode head) {ListNode fast = head;ListNode slow = head;while(fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;if(fast == slow) {return true;}}if(fast == null || fast.next == null) {return false;}return true;}}
10.环形链表II
链接 142. 环形链表 II – 力扣(LeetCode)
题目要求
分析一下
上代码
public class Solution {public ListNode detectCycle(ListNode head) {ListNode fast = head;ListNode slow = head;while(fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;if(fast == slow) {break;}}if(fast == null || fast.next == null) { return null;}slow = head;while(slow != fast) {fast = fast.next;slow = slow.next;}return fast;}}