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203.移除链表元素

若要移除链表中某一个结点，需要分为两种情况：
1.移除中间结点：让上一个结点next指向下一个结点
2.移除头结点，则直接让头结点后移即可
3.因此，为了想让链表移除结点操作逻辑统一，则可以加上一个虚拟头结点，也就是让虚拟头结点dummyhead的 next 指向头结点head。那么这样的话，移除头结点也只需要将虚拟头结点指向头结点的下一个结点就行了，逻辑得到统一。
4.而且，一般链表类题目的返回值是链表的头结点，题目中难以去额外保存一个头结点。如果设置了虚拟头结点的话，那么就可以直接返回dummyhead.next

/**
 * 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 head;
        }
        ListNode dummyNode = new ListNode(0);
        dummyNode.next = head;
        ListNode pre = dummyNode;
        ListNode cur = head;

        while(cur != null){

            if(cur.val == val){
                pre.next = cur.next;
            }else{
                pre = cur;
            }
            cur = cur.next;
        }
        return dummyNode.next;
    }
}

707.设计链表

题目链接: 设计链表

设计链表逻辑比较多，主要是一些index边界条件需要额外判判断。此处我设计的是单链表

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 {

    int size;
    ListNode dummyhead;

    public MyLinkedList() {
        size = 0;
        dummyhead = new ListNode(0);
    }
    
    public int get(int index) {
        if(index < 0 || index >= size){
            return -1;
        }
        ListNode currentNode = dummyhead;
        for(int i = 0; i <= index; i++){
            currentNode = currentNode.next;
        }
        return currentNode.val;
    }
    
    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;
        }
        ListNode preNode = dummyhead;
        for(int i = 0; i < index; i++){
            preNode = preNode.next;
        }
        ListNode addNode = new ListNode(val);
        addNode.next = preNode.next;
        preNode.next = addNode;
        size++;
    }
    
    public void deleteAtIndex(int index) {
        if (index < 0 || index >= size) {
            return;
        }
       
        ListNode preNode = dummyhead;
        for(int i = 0; i < index; i++){
            preNode = preNode.next;
        }
        preNode.next = preNode.next.next;
        size--;
    }
}

/**
 * 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

题目链接: 反转链表

不要去新建一个链表，而只需要在原来的链表上操作
定义pre cur temp 三个ListNode，实现链表反转

/**
 * 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 pre = null;
        ListNode cur = head;
        ListNode temp = null;

        while(cur != null){
            temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}