206. 反转链表 - 力扣（LeetCode）

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode reverseList(ListNode head) {ListNode cur = head;ListNode pre = null;ListNode next =  null;while (cur != null) {next = cur.next;cur.next = pre;pre = cur;cur = next;}return pre;}
}

92. 反转链表 II - 力扣（LeetCode）

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode reverseBetween(ListNode head, int left, int right) {ListNode dummy = new ListNode(0);dummy.next = head;ListNode p0 = dummy;for (int i = 0; i < left - 1; i++) {p0 = p0.next;}ListNode pre = null;ListNode next = null;ListNode cur = p0.next;for (int i = 0; i < right - left + 1; i++) {next = cur.next;cur.next = pre;pre = cur;cur = next;}p0.next.next = cur;p0.next = pre;return dummy.next;}
}

25. K 个一组翻转链表 - 力扣（LeetCode）

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode reverseKGroup(ListNode head, int k) {int n = 0;ListNode cur = head;while (cur != null) {cur = cur.next;n++;}ListNode dummy = new ListNode(0);dummy.next = head;ListNode p0 = dummy;cur = p0.next;ListNode pre = null;ListNode next = null;while (n >= k) {n -= k;for (int i = 0; i < k; i++) {next = cur.next;cur.next = pre;pre = cur;cur = next;}next = p0.next;p0.next.next = cur;p0.next = pre;p0 = next;}return dummy.next;}
}

课后作业：

24. 两两交换链表中的节点 - 力扣（LeetCode）

延续上面的写法，创建哨兵节点p0，和反转K个group代码一模一样，就是k==2

time:O(n) 

space:O(1)

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode swapPairs(ListNode head) {int n = 0;ListNode cur = head;while (cur != null) {cur = cur.next;n++;}ListNode dummy = new ListNode(0);dummy.next = head;ListNode p0 = dummy;cur = p0.next;ListNode pre = null;ListNode next = null;while (n >= 2) {n -= 2;for (int i = 0; i < 2; i++) {next = cur.next;cur.next = pre;pre = cur;cur = next;}next = p0.next;p0.next.next = cur;p0.next = pre;p0 = next;}return dummy.next;}
}

方法二：递归,感觉容易想一点

time: O(n),

space:O(n),递归需要 O(n)\mathcal{O}(n)O(n) 的栈空间

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode swapPairs(ListNode head) {if (head == null || head.next == null) return head;ListNode newHead = head.next;ListNode node = swapPairs(newHead.next);newHead.next = head;head.next = node;return newHead;}
}

445. 两数相加 II - 力扣（LeetCode）

方法1：可以用递归，也可用迭代

1.反转链表1

2.反转链表2

3.addTwo得到链表3

4.反转链表3

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {l1 = reverse(l1);l2 = reverse(l2);ListNode l3 = addTwo(l1, l2);return reverse(l3);}private ListNode reverse(ListNode head) {if (head == null || head.next == null) return head;ListNode newHead = reverse(head.next);head.next.next = head;head.next = null;return newHead;}private ListNode addTwo(ListNode l1, ListNode l2) {ListNode dummy = new ListNode(0);ListNode cur = dummy;int carry = 0;int sum = 0;while (l1 != null && l2 != null) {sum = l1.val + l2.val + carry;carry = sum / 10;cur.next = new ListNode(sum % 10);cur = cur.next;l1 = l1.next;l2 = l2.next;}while (l1 != null) {sum = l1.val + carry;carry = sum / 10;cur.next = new ListNode(sum % 10);cur = cur.next;l1 = l1.next;}while (l2 != null) {sum = l2.val + carry;carry = sum / 10;cur.next = new ListNode(sum % 10);cur = cur.next;l2 = l2.next;}if (carry != 0) {cur.next = new ListNode(carry);}return dummy.next;}
}

方法2：用stack

两个stack，其他和数组做法没啥区别

2816. 翻倍以链表形式表示的数字 - 力扣（LeetCode）

直接用445的代码就行

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode doubleIt(ListNode head) {head = reverse(head);ListNode res = addTwo(head, head);return reverse(res);}private ListNode reverse(ListNode head) {if (head == null || head.next == null) return head;ListNode newHead = reverse(head.next);head.next.next = head;head.next = null;return newHead;}private ListNode addTwo(ListNode l1, ListNode l2) {ListNode dummy = new ListNode(0);ListNode cur = dummy;int carry = 0;int sum = 0;while (l1 != null && l2 != null) {sum = l1.val + l2.val + carry;carry = sum / 10;cur.next = new ListNode(sum % 10);cur = cur.next;l1 = l1.next;l2 = l2.next;}while (l1 != null) {sum = l1.val + carry;carry = sum / 10;cur.next = new ListNode(sum % 10);cur = cur.next;l1 = l1.next;}while (l2 != null) {sum = l2.val + carry;carry = sum / 10;cur.next = new ListNode(sum % 10);cur = cur.next;l2 = l2.next;}if (carry != 0) {cur.next = new ListNode(carry);}return dummy.next;}
}