代码随想录DAY03| 203.移除链表元素、707.设计链表、206.反转链表

1. 链表的定义

public class ListNode {
    // 结点的值
    int val;

    // 下一个结点
    ListNode next;

    // 节点的构造函数(无参)
    public ListNode() {
    }

    // 节点的构造函数(有一个参数)
    public ListNode(int val) {
        this.val = val;
    }

    // 节点的构造函数(有两个参数)
    public ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}

力扣题目链接：移除链表元素

有三种解法：

1. 遍历链表并删除，这样需要单独讨论头节点，这就引出解法2
2. 虚拟头节点，在头节点前加一个，这样就可以统一操作
3. 递归，对节点进行递归讨论

/**
 * 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; }
 * }
 */
 //解法一
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        while(head != null && head.val == val){
            head = head.next;
        }
        ListNode prev = head;
        while(prev != null && prev.next != null){
            if(prev.next.val == val){
                prev.next = prev.next.next;
            }else{
                prev = prev.next;
            }
        }

        return head;
    }
}

//解法二
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode();
        dummy.next = head;
        ListNode pre = dummy;
        while(pre != null && pre.next != null){
            if(pre.next.val == val){
                pre.next = pre.next.next;
            }else{
                pre = pre.next;
            }
        }

        return dummy.next;
    }
}

//方法三
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if(head == null){
            return head;
        }else if(head.val == val){
            return removeElements(head.next,val);
        }else{
            head.next = removeElements(head.next,val);
            return head;
        }
    }
}

2. 设计链表

注意使用虚拟头节点。

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 MyLinkedList {

    int size;
    //此处的head是虚拟头节点，方便后续操作
    ListNode head;

    public MyLinkedList() {
        size = 0;
        head = new ListNode(0);
    }
    
    public int get(int index) {
        if(index < 0 || index >= size){
            return -1;
        }
        ListNode currentNode = head;
        for(int i = 0; i <= index;i++){
            currentNode = currentNode.next;
        }
        return currentNode.val;
    }
    
    public void addAtHead(int val) {
        ListNode node = new ListNode(val);
        node.next = head.next;
        head.next = node;
        size++;
    }
    
    public void addAtTail(int val) {
        ListNode currentNode = head;
        ListNode node = new ListNode(val);
        while(currentNode.next != null){
            currentNode = currentNode.next;
        }
        currentNode.next = node;
        size++;
    }
    
    public void addAtIndex(int index, int val) {
        if(index > size){
            return;
        }
        if(index < 0){
            index = 0;
        }
        ListNode node = new ListNode(val);
        ListNode pre = head;
        for(int i = 0;i < index; i++){
            pre = pre.next;
        }
        node.next = pre.next;
        pre.next = node;
        size++;
    }
    
    public void deleteAtIndex(int index) {
        if(index < 0 || index >= size){
            return;
        }
        ListNode pre = head;
        for(int i = 0;i < index; i++){
            pre = pre.next;
        }
        pre.next = pre.next.next;
        size--;
    }
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList obj = new MyLinkedList();
 * int param_1 = obj.get(index);
 * obj.addAtHead(val);
 * obj.addAtTail(val);
 * obj.addAtIndex(index,val);
 * obj.deleteAtIndex(index);
 */

3. 反转链表

1. 注意pre节点不需要连上next
2. 用递归的思想。出口是cur == null，每次的返回值并没有被用到，要层层返回给调用处。

//解法一
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode cur = head;
        ListNode temp = null;
        ListNode pre = null;
        while(cur != null){
            temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }

        return pre;
    }
}

class Solution {
    public ListNode reverseList(ListNode head) {
        return reverse(null, head);
    }

    public ListNode reverse(ListNode pre, ListNode cur){
        if(cur == null){
            return pre;
        }
        ListNode temp = new ListNode();
        temp = cur.next;
        cur.next = pre;
        return reverse(cur, temp);
    }
}