简介

记录一下自己刷题的历程以及代码。写题过程中参考了 代码随想录。会附上一些个人的思路，如果有错误，可以在评论区提醒一下。

[简单] 203. 移除链表元素

原题链接
一前一后两个指针做遍历操作，记得 开头对空链表做判断，结尾对head是否需要删除做判断

class Solution {public ListNode removeElements(ListNode head, int val) {if (head == null) {return head;}ListNode before = head; // 标记删除元素前一个元素ListNode del = head.next; // 标记删除元素while(del != null){// 删除if(del.val == val) {before.next = del.next;del = before.next;}else{// 前进before = del;del = del.next;}}if(head.val == val) return head.next;return head;}
}

[中等] 707. 设计链表

原题链接
根据平时写Java工程的习惯加入了length属性记录链表长度，方便做过界判断。
链表设计上默认有一个虚的头结点，不计算长度，方便做插入和删除操作。

class MyLinkedList {//考虑存在虚的头结点MyLinkedList next;int val;//维护一个私有变量length表示长度//方便查找时做越界判断private int length;public MyLinkedList() {}public int get(int index) {if(index > length - 1) return -1;int count = index;MyLinkedList pointer = this.next;while(count != 0){pointer = pointer.next;count--;}return pointer.val;}public void addAtHead(int val) {MyLinkedList newHead = new MyLinkedList();newHead.val = val;newHead.next = this.next;this.next = newHead;length++;}public void addAtTail(int val) {MyLinkedList newTail = new MyLinkedList();newTail.val = val;MyLinkedList pointer = this;while(pointer.next != null) pointer = pointer.next;pointer.next = newTail;length++;}public void addAtIndex(int index, int val) {if(index > length) return;else if(index == 0){addAtHead(val);return;}else if(index == length) {addAtTail(val);return;}//以上情况之外，插入的位置都会有前置节点int count = index;MyLinkedList pointer = this.next;while(count > 1){pointer = pointer.next;count--;}MyLinkedList newNode = new MyLinkedList();newNode.val = val;newNode.next = pointer.next;pointer.next = newNode;length++;}public void deleteAtIndex(int index) {//头跟尾分开考虑if(index > length - 1) return;int count = index;MyLinkedList pointer = this;while(count > 0){pointer = pointer.next;count--;}if(pointer.next != null){pointer.next = pointer.next.next;}length--;}public void printList(){System.out.print("[");System.out.print(val);MyLinkedList pointer = this.next;while(pointer != null){System.out.print("," + pointer.val);pointer = pointer.next;}System.out.print("]");System.out.print("    length: " + length);System.out.print("\n");}
}public class main {public static void main(String[] args) {MyLinkedList myLinkedList = new MyLinkedList();myLinkedList.printList();myLinkedList.addAtHead(1);myLinkedList.printList();myLinkedList.addAtTail(3);myLinkedList.printList();myLinkedList.addAtIndex(1, 2);myLinkedList.printList();System.out.println(myLinkedList.get(1));myLinkedList.deleteAtIndex(0);myLinkedList.printList();System.out.println(myLinkedList.get(0));}
}

[简单] 206. 反转链表

原题链接

直接在原链表上挨个改变next指针指向做原地倒置。注意边界情况的判断即可

class Solution {public ListNode reverseList(ListNode head) {if(head == null) return head;ListNode pre = null;ListNode cur = head;ListNode temp = head.next;while(cur != pre){cur.next = pre;pre = cur;if(temp != null) {cur = temp;temp = cur.next;}}return cur;}
}

在有虚头结点的情况下可以做头插法反转，同理这题也可以自己给出一个虚头结点使用头插法反转链表。

class Solution {public ListNode reverseList(ListNode head) {ListNode vHead = new ListNode();vHead.next = null;ListNode cur = head;while(cur != null){ListNode temp = cur.next;cur.next = vHead.next;vHead.next = cur;cur = temp;}return vHead.next;}
}

[简单] 24. 两两交换链表中的节点

原题链接

定义一个虚的头结点方便后续循环操作，因为转换两个节点不只是这两个节点的next需要做处理，两个节点的上一个节点的next也需要重新改变指向，缺少虚头结点的情况下，每次在head上做转换操作的时候不需要操作前置节点，其他情况都需要，就无法统一操作。

class Solution {public ListNode swapPairs(ListNode head) {if (head == null || head.next == null) return head; //长度为0或1直接返回ListNode vHead = new ListNode();vHead.next = head; //设定虚头结点ListNode pre = vHead;ListNode cur = head;while(cur != null && cur.next != null){pre.next = cur.next;cur.next = pre.next.next;pre.next.next = cur;pre = cur;cur = cur.next;}return vHead.next;}
}

[简单] 19. 删除链表的倒数第 N 个结点

原题链接

删除一个节点需要找到他的前置节点，同样设置虚头结点方便操作（没有虚头结点就是多一个对删除第一个元素的判断，因为除了第一个元素之外其他元素都有前置节点需要改变next指针）。使用快慢指针的思路，让前一个指针先走n步，之后两个指针一起前进，第二个指针就会比第一个指针慢n步，就能指向我们需要删除的节点。

class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode vHead = new ListNode();vHead.next = head;ListNode node = vHead;ListNode delete = vHead;while(n-- != 0){node = node.next;}while(node.next != null){node = node.next;delete = delete.next;}delete.next = delete.next.next;return vHead.next;}
}