代码随想录训练营二刷第三天 | 203.移除链表元素 707.设计链表 206.反转链表

一、203.移除链表元素

题目链接：https://leetcode.cn/problems/remove-linked-list-elements/
思路：使用虚拟头结点，两个指针，一个是遍历指针，一个是指向当前位置的上一个位置

/**
 * 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) {
        if (head == null) return null;
        ListNode node = new ListNode(-1, head);
        ListNode pre = node, cur = head;
        while (cur != null) {
            if (cur.val == val) {
                pre.next = cur.next;
            }else {
                pre = cur;
            }
            cur = cur.next;
        }
        return node.next;
    }
}

一、707.设计链表

题目链接：https://leetcode.cn/problems/design-linked-list/
思路：需要定义单独一个链表对象，然后用MyLinkedList来操作定义的链表对象，默认初始化一个虚拟头结点。

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 len ;
    ListNode head;
    public MyLinkedList() {
        len = 0;
        head = new ListNode(-1);
    }

    public int get(int index) {
        if (index < 0 || index >= len) {
            return -1;
        }
        int i = 0;
        ListNode p = head;
        while (i <= index) {
            i++;
            p = p.next;
        }
        return p.val;
    }

    public void addAtHead(int val) {
        addAtIndex(0, val);
    }

    public void addAtTail(int val) {
        addAtIndex(len, val);
    }

    public void addAtIndex(int index, int val) {
        if (index > len) {
            return;
        }
        if (index < 0) {
            index = 0;
        }
        int i = 0;
        ListNode p = head;
        while (i < index) {
            i++;
            p = p.next;
        }
        p.next = new ListNode(val, p.next);
        len++;
    }

    public void deleteAtIndex(int index) {
        if (index < 0 || index >= len) {
            return;
        }
        ListNode pre = head, cur = head.next;
        int i = 0;
        while (i < index) {
            i++;
            pre = cur;
            cur = cur.next;
        }
        pre.next = cur.next;
        len--;
    }
}

/**
 * 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);
 */

一、206.反转链表

题目链接：https://leetcode.cn/problems/reverse-linked-list/
思路：很经典的题目，每次操作前用一个临时指针保存cur的next，然后让cur 指向pre，只有pre跟进一步，pre=cur，然后cur=temp

/**
 * 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 || head.next == null) {
            return head;
        }
        ListNode pre = null, cur = head, temp = null;
        while (cur != null) {
            temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}