算法训练（leetcode）二刷第三天 | 203. 移除链表元素、707. 设计链表、206. 反转链表

203. 移除链表元素

leetcode题目地址

思路二

创建虚节点作为头结点指向原链表，对链表中的目标元素进行删除，最后返回虚节点的下一个结点。

时间复杂度： $O(n)$
空间复杂度： $O(1)$

// java
/**
 * 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 removeElements(ListNode head, int val) {
        ListNode dummyHead = new ListNode(-1, head);
        // dummyHead.next = head;

        ListNode p = dummyHead;

        while(p.next != null){
            if(p.next.val == val) p.next = p.next.next;
            else p = p.next;
        }
        return dummyHead.next;
    }
}

思路二

创建虚节点作为一条新链表的头结点，next为空。从原链表中寻找不是目标值的结点，接到新链表尾。

时间复杂度： $O(n)$
空间复杂度： $O(1)$

/**
 * 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 removeElements(ListNode head, int val) {
        ListNode newHead = new ListNode(-1, null), q = newHead;
        ListNode p = head, pre = null;
        while(p != null){
            if(p.val != val){
            	// 记录节点地址
                pre = p;
                // 指针后移（防止修改next断链）
                p = p.next;
                // 将该节点接入新链表
                pre.next = q.next;
               	q.next = pre;
                q = pre;
            }else{
                p = p.next;
            }
        }
        return newHead.next;
    }
}

707. 设计链表

leetcode题目地址

维护头指针、尾指针以及链表长度三个属性，这样可以避免在尾部插入元素时遍历整个链表。

时间复杂度： 涉及index的操作为 $O(index)$, 其余操作为$O(1)$
空间复杂度： $O(n)$

// java
class Node{
    int val;
    Node next;
    public Node(){
        next = null;
    }
    public Node(int val){
        this.val = val;
        next = null;
    }
}

class MyLinkedList {

    private Node head;
    private Node tail;
    private int length;

    public MyLinkedList() {
        this.head = new Node();
        this.tail = head;
        this.length = 0;
    }
    
    public int get(int index) {
        if(index < 0 || index >= this.length) return -1;
        Node p = head.next;
        while(index--!=0) p = p.next;
        return p.val;
    }   
    
    public void addAtHead(int val) {
        Node newNode = new Node(val);
        newNode.next = this.head.next;
        this.head.next = newNode;
        if(tail.equals(head)) tail = newNode;
        this.length++;
    }
    
    public void addAtTail(int val) {
        Node newNode = new Node(val);
        newNode.next = this.tail.next;
        this.tail.next = newNode;
        tail = newNode;
        this.length++;
    }
    
    public void addAtIndex(int index, int val) {
        if(index < 0 || index > this.length) return;
        if(index == this.length) this.addAtTail(val);
        else{
            Node newNode = new Node(val);
            Node pre = head;
            while(index--!=0){
                pre = pre.next;
            }
            newNode.next = pre.next;
            pre.next = newNode;
            this.length++;
        }
    }
    
    public void deleteAtIndex(int index) {
        if(index < 0 || index >= this.length) return;
        Node pre = head;
        while(index--!=0){
            pre = pre.next;
        }
        if(this.tail.equals(pre.next)){
            tail = pre;
        }
        pre.next = pre.next.next;
        this.length--;
    }
}

/**
 * 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);
 */

206. 反转链表

leetcode题目地址

使用头插法实现单链表就地逆置。

时间复杂度： $O(n)$
空间复杂度： $O(1)$

// java
/**
 * 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 dummyHead = new ListNode(0, null);
        ListNode p = head, cur=null;
        while( p != null){
            cur = p;
            p = p.next;

            cur.next = dummyHead.next;
            dummyHead.next = cur;
        } 
        return dummyHead.next;
    }
}