代码随想录day3 Java版

今天开始搞链表的题目，因为用的Java，先看了下力扣提供的链表：

/**
 * 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; }
 * }
 */

一定注意拿到指针后每次判断都要看看现在是不是空指针，好像很容易踩坑

203.移除链表元素

先自己写了个常规版本，分类讨论做法，对于头节点特殊处理

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        while (head != null && head.val == val) {
            head = head.next;
        }
        if (head == null) return head;
 
        ListNode p = head;
        while (p != null) {
            if (p.next != null && p.next.val == val) {
                p.next = p.next.next;
            } else {
                p = p.next;
            }
            
        }
        return head;
    }
}

看了网上经常说的哨兵节点，加入后就可以消除头节点的特殊性

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) return head;
 
        ListNode dummy = new ListNode(-1,head);
        ListNode p = dummy;
        while(p != null) {
            if (p.next != null && p.next.val == val) p.next = p.next.next;
            else p = p.next;
        }
        return dummy.next;
    }
}

707.设计链表

不用解释，直接手搓就行

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

206.反转链表

老题了，重点在于反向后链表会断裂，所以必须提前保存原来的下一个才能继续遍历

class Solution {
    public ListNode reverseList(ListNode head) {
        //本质上就是箭头的转换
        ListNode cur = head, pre = null, temp = null;
        while (cur != null) {
            temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}