代码随想录四刷day3

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前言

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双指针的经典应用，如果要删除倒数第n个节点，让fast移动n步，然后让fast和slow同时移动，直到fast指向链表末尾。删掉slow所指向的节点就可以了。

一、力扣707. 设计链表

class MyLinkedList {
    private Node head;
    private int count;
    class Node {
        int val;
        Node next;
        public Node(int val, Node next){
            this.val = val;
            this.next = next;
        }
        public Node(){
        }
    }

    public MyLinkedList() {
        this.head = new Node(-1,null);
        this.count = 0;
    }
    
    public int get(int index) {
        if(index >= count){
            return -1;
        }
        Node p = head;
        for(int i = 0; i <= index; i ++){
            p = p.next;
        }
        return p.val;
    }
    
    public void addAtHead(int val) {
        addAtIndex(0,val);
    }
    
    public void addAtTail(int val) {
        addAtIndex(count,val);
    }
    
    public void addAtIndex(int index, int val) {
        if(index > count){
            return;
        }
        Node pre = this.head;
        Node p = new Node(val,null);
        while(index > 0){
            index --;
            pre = pre.next;
        }
        p.next = pre.next;
        pre.next = p;
        count ++;
    }
    
    public void deleteAtIndex(int index) {
        if(index >= count){
            return;
        }
        Node pre = head;
        for(int i = 0; i < index; i ++){
            pre = pre.next;
        }
        pre.next = pre.next.next;
        count --;
    }
}

/**
 * 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);
 */

二、力扣206. 反转链表

/**
 * 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) {
        if(head == null){
            return head;
        }
        ListNode L = new ListNode(-1,null);
        ListNode p = head;
        while(p != null){
            ListNode s = p;
            p = p.next;
            s.next = L.next;
            L.next = s;
        }
        return L.next;
    }
}

三、力扣24. 两两交换链表中的节点

/**
 * 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 p1 = head, p2 = head.next;
        ListNode L = new ListNode(-1,head);
        ListNode pre = L, s1, s2;
        while(p1 != null && p2 != null){
            s1 = p1; s2 = p2;
            if(p2.next == null){
                p1 = null;
                p2 = null;
            }else if(p2.next != null && p2.next.next == null){
                p1 = p2.next;
                p2 = null;
            }else if(p2.next != null && p2.next.next != null){
                p1 = p2.next;
                p2 = p1.next;
            }
            pre.next = s2;
            s2.next = s1;
            pre = s1;
            s1.next = p1;
        }
        return L.next;
    }
}

四 、力扣19. 删除链表的倒数第 N 个结点

/**
 * 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 removeNthFromEnd(ListNode head, int n) {
        ListNode L = new ListNode(-1,head);
        ListNode pre = L, p = L;
        for(int i = 0; i <= n; i ++){
            p = p.next;
        }
        while(p != null){
            p = p.next;
            pre = pre.next;
        }
        pre.next = pre.next.next;
        return L.next;
    }
}