题目一：

203. 移除链表元素 - 力扣（LeetCode）

public ListNode removeElements(ListNode head, int val) {//1. 如果链表为null，直接返回headif (head == null) {return head;}//2. 定义快慢指针ListNode pre = head;ListNode del = pre.next;while (pre.next != null) {if (del.val == val) {pre.next = del.next;//del = del.next;} else {pre = pre.next;//del = del.next;}del = del.next;}//3. 排查头节点if (head.val == val) {head = head.next;}//4. 返回return head;}

题目二：

206. 反转链表 - 力扣（LeetCode）

public ListNode reverseList(ListNode head) {//1. 如果链表为null，直接返回headif (head == null) {return head;}//2. 正常ListNode cur = head.next;//待交换节点//3. 将头节点的next置为nullhead.next = null;//4. 进行交换while (cur != null) {//5. 记录待交换节点的下一个节点ListNode curN = cur.next;cur.next = head;head = cur;cur = curN;}//5. 返回return head;}

题目三：

876. 链表的中间结点 - 力扣（LeetCode）

public ListNode middleNode(ListNode head) {//1. 题目提示：链表一定不为null//2. 定义快慢指针ListNode fast = head;//每次走两步ListNode slow = head;//每次走一步while (fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;}//3. 返回return slow;}

题目四：正

面试题 02.02. 返回倒数第 k 个节点 - 力扣（LeetCode）

public int kthToLast(ListNode head, int k) {//新增条件：k不一定合法//1. 如果链表为null，k<=0if (head == null || k <= 0) {return -1;}//2. 定义快慢指针ListNode fast = head;ListNode slow = head;//3. 快指针先走k-1步,同时判断k的合法性for (int i = 0; i < k - 1; i++) {//【解析】：当fast.next == null时，fast不能再继续走了//if语句写在前面的原因，如果fast.next == null，往后走了之后//fast == null，再进入if语句会造成空指针异常if (fast.next == null) {return -1;}fast = fast.next;}//4. 快慢指针同时走while (fast.next != null) {fast = fast.next;slow = slow.next;}//5. 返回return slow.val;}

题目五：正

21. 合并两个有序链表 - 力扣（LeetCode）

public ListNode mergeTwoLists(ListNode head1, ListNode head2) {//1. 定义傀儡节点ListNode head = new ListNode(-1);ListNode cur = head;//2. 开始拼接while (head1 != null && head2 != null) {if (head1.val < head2.val) {cur.next = head1;head1 = head1.next;} else {cur.next = head2;head2 = head2.next;}cur = cur.next;}//3. 根据出循环的原因分别作出反应（即使两个链表都为空也没关系）if (head1 == null) {cur.next = head2;}if (head2 == null) {cur.next = head1;}//3. 返回return head.next;}

题目六：

链表分割_牛客题霸_牛客网 (nowcoder.com)

public ListNode partition(ListNode pHead, int x) {// write code here//1. 如果链表为nullif (pHead == null) {return pHead;}//2. 定义两组傀儡节点ListNode head1 = new ListNode(-1);ListNode cur1 = head1;ListNode head2 = new ListNode(-1);ListNode cur2 = head2;//3. 遍历所给链表while (pHead != null) {if (pHead.val < x) {cur1.next = pHead;cur1 = cur1.next;} else {cur2.next = pHead;cur2 = cur2.next;}pHead = pHead.next;}//4. 合并两个傀儡链表//① 如果前一个链表为nullif (head1.next == null) {return head2.next;}//② 如果前一个链表不为null（包含后一个链表为null和不为null的情况）cur1.next = head2.next;cur2.next = null;//最后一个节点的值给了前一个链表，则此时后一个链表的最后一个节点的next一定不是nullreturn head1.next;}

题目七：正

链表的回文结构_牛客题霸_牛客网 (nowcoder.com)

public boolean chkPalindrome(ListNode A) {// write code here//1. 如果链表为nullif (A == null) {return true;}//2. 找到中间节点ListNode fast = A;//每次走两步ListNode slow = A;//每次走一步while (fast != null && fast.next != null) {fast = fast.next.next;slow = slow.next;}//3. 翻转中间节点之后的所有节点ListNode cur = slow.next;//待翻转节点while (cur != null) {ListNode curN = cur.next;cur.next = slow;slow = cur;cur = curN;}//4. 判断是否为回文fast = A;while (fast != slow) {//链表节点个数为奇数，fast==slow为结束标志if (fast.val != slow.val) {//不对称return false;} else {//对称if (fast.next == slow) {//链表节点个数为偶数，fast.next==slow为结束标志return true;//如果是对称的条件，并且满足了偶数节点个数的结束条件，则可直接返回}fast = fast.next;slow = slow.next;}}//5. 返回return true;}

题目八：正

160. 相交链表 - 力扣（LeetCode）

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {//题目已经提示两个链表的节点个数一定不为0//1. 计算两个链表的长度,并假设链表A更长int countA = 0;int countB = 0;ListNode curA = headA;ListNode curB = headB;while (curA != null) {countA++;curA = curA.next;}while (curB != null) {countB++;curB = curB.next;}//2. 判断究竟哪个链表更长int sub = countA - countB;curA = headA;curB = headB;if (sub < 0) {//此时B链表更长curA = headB;curB = headA;sub = countB - countA;}//3. 让更长的链表先走sub步for (int i = 0; i < sub; i++) {curA = curA.next;}//4. 找相遇节点while (curA != null) {//只用判断curA和curB中是否有一个为null即可if (curA == curB) {return curA;}curA = curA.next;curB = curB.next;}//5. 代码走到这里一定是因为两个链表不相交return null;}

题目九：正

141. 环形链表 - 力扣（LeetCode）

public boolean hasCycle(ListNode head) {//定义快慢指针（下列代码链表为null也能正确处理）ListNode fast = head;//一次走两步ListNode slow = head;//一次走一步while (fast != null && fast.next != null) {//有顺序要求的原因，如果链表为空，后面的会空指针异常fast = fast.next.next;slow = slow.next;if (fast == slow) {return true;}}return false;}

题目十：正

142. 环形链表 II - 力扣（LeetCode）

public ListNode detectCycle(ListNode head) {//1. 找到快慢指针相遇的节点//定义快慢指针（下列代码链表为null也能正确处理）ListNode fast = head;//一次走两步ListNode slow = head;//一次走一步while (fast != null && fast.next != null) {//有顺序要求的原因，如果链表为空，后面的会空指针异常fast = fast.next.next;slow = slow.next;if (fast == slow) {break;}}//2. 判断是如何出while循环的//关于不能以fast != slow作为循环条件的原因//答：当链表为null,或链表只有一个节点时，也满足该条件，//此时则不能判定是fast==slow的情况if (fast == null || fast.next == null) {return null;}//3. 寻找入环点fast = head;while (fast != slow) {fast = fast.next;slow = slow.next;}//4. 返回return fast;}