【链表专题】

移除链表元素

• 移除链表元素（Leetcode:203）
  给你一个链表的头节点 head 和一个整数 val ，请你删除链表中所有满足 Node.val == val 的节点，并返回 新的头节点 。
  

/**
 * Definition for singly-linked list.//注意基本的链表定义方法
 * 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) {
        //非空判断
        if (head == null) {
            return null;
        }
        //定义虚拟头节点
        ListNode dummy = new ListNode(-1, head);
        ListNode pre = dummy;
        ListNode cur = head;

        //删除指定元素的值节点
        while(cur != null) {
            if (cur.val == val) {
                pre.next = cur.next;
            } else {
                pre = cur;
            }
            cur = cur.next;
        }
        //返回新的链表的头节点
        return dummy.next;
    }
}
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设计链表

• 设计链表（Leetcode:707）
  

class ListNode {
    int val;
    ListNode next;

    public ListNode() {

    }

    public ListNode(int val) {
        this.val = val;
    }

    public ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }

}

class MyLinkedList {

    //size存储链表元素的个数
    int size;
    //虚拟头节点
    ListNode head;

    //初始化链表
    public MyLinkedList() {
        size = 0;
        head = new ListNode(0);
    }
    
    //获取第index个节点的数值，注意index是从0开始的，第0个节点就是头结点
    public int get(int index) {
        if (index < 0 || index >= size) {
            return -1;
        }
        //当前指针先指向虚拟头节点
        ListNode cur = head;

        for (int i = 0; i <= index; i++) {
            cur = cur.next;
        }

        return cur.val;
    }
    
     //在链表最前面插入一个节点，等价于在第0个元素前添加
    public void addAtHead(int val) {
        addAtIndex(0, val);
    }
    
    public void addAtTail(int val) {
        addAtIndex(size, val);
    }
    
    public void addAtIndex(int index, int val) {
        if (index > size) {
            return;
        }

        if (index < 0) {
            index = 0;
        }

        size++;

        ListNode pre = head;
        
        for (int i = 0; i < index; i++) {//举个极端的在末尾插入的例子，如数组[1]，在index为1处插入元素，循环结束会出现pre指向null的情况，导致引用失效，所以i不能=index
            pre = pre.next;
        }

        ListNode toAdd = new ListNode(val);
        toAdd.next = pre.next;
        pre.next = toAdd;
    }
    
    public void deleteAtIndex(int index) {
        if (index < 0 || index >= size) {
            return;
        }

        size--;

        if (index == 0) {
            head = head.next;
            return;
        }

        ListNode pre = head;
        for (int i = 0; i < index; i++) {//极端例子，考虑到pre和pre.next都不能为空；如[1,2,3]删除index为2的元素，
            pre = pre.next;
        }

        pre.next = pre.next.next;
    }
}

/**
 * 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);
 */
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反转链表

• 反转链表（Leetcode:206）

/**
 * 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) {
        //双指针法：先定义两个指针pre和cur分别指向null和head,一个临时指针temp
        ListNode pre = null;
        ListNode cur = head;
        ListNode temp = null;

        while (cur != null) {//循环遍历
            //临时指针先存放cur.next,因为它等会会改变方向，避免丢失引用
            temp = cur.next;
            //改变指针指向，由指向后一个节点改为指向前一个节点
            cur.next = pre;
            //移动pre, cur向后移，继续后续的反转的过程
            pre = cur;
            cur = temp;
        }

        return pre;
        
    }
}
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