链表篇（上 代码随想录二刷+总结）

1.移除链表元素

leetcode 203

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode newHead = new ListNode(0,head);
        ListNode pre = newHead;
        while(head!=null){
            if(head.val==val){
                pre.next = head.next;
                head = head.next;
            }else{
                pre = head;
                head = head.next;
            }
        }
        return newHead.next;
    }
}

链表问题一般要考虑构建虚拟节点，让整个链表可以具有统一的操作。这道题就是最基本操作，只做太久没做链表题的复习。

2.设计链表

leetcode 707

class MyLinkedList {
    class Node{
        int val;
        Node next;
        Node(int val){
            this.val = val;
        }
    }
    int length;
    Node head;
    public MyLinkedList() {
        //初始化
        head = null;
        length=0;
    }
    
    public int get(int index) {
        if(index>=length||index<0){
            return -1;
        }
        int i=0;
        Node getNode = head;
        while(i<index){
            getNode = getNode.next;
            i++;
        }
        return getNode.val;
    }
    
    public void addAtHead(int val) {
        Node newHead = new Node(val);
        newHead.next = head;
        head = newHead;
        length++;
    }
    
    public void addAtTail(int val) {
        Node MyNode = new Node(val);
        MyNode.next = null;
        if(length == 0){
            head = MyNode;
        }else{
            Node pre = head;
            while(pre.next!=null){
                pre = pre.next;
            }
            pre.next = MyNode;
        }
        length++;
    }
    
    public void addAtIndex(int index, int val) {
        if(index>length){
            return;
        }
        if(index<=0){
            addAtHead(val);
            return;
        }
        if(index == length){
            addAtTail(val);
            return;
        }
        Node temp = head;
        int i = 0;
        while(i<index-1){
            temp = temp.next;
            i++;
        }
        Node addNode = new Node(val);
        addNode.next = temp.next;
        temp.next = addNode;
        length++;
    }
    
    public void deleteAtIndex(int index) {
        if(index==0){
            head = head.next;
            length--;
        }else if(index>0&&index<length){
            Node temp = head;
            for(int i=0;i<index-1;i++){
                temp = temp.next;
            }
            temp.next = temp.next.next;
            length--;
        }
    }

}

练手复习即可

3.反转链表

leetcode 206

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode NEXT;
        ListNode cur = head;
        while(cur!=null){
            NEXT = cur.next;
            cur.next = pre;
            pre = cur;
            cur = NEXT;
        }
        return pre;
    }
}

因为一个节点反转后无法找到正序的下一个节点，所以反转过程前记录下一个节点。

4.两两交换链表中的节点

leetcode 24

class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode cur = new ListNode(-1,head);
        ListNode res = cur;
        while(cur.next!=null&&cur.next.next!=null){
            ListNode temp1 = cur.next.next.next;
            ListNode temp2 = cur.next;
            cur.next = cur.next.next;
            cur.next.next = temp2;
            cur.next.next.next = temp1;
            cur = cur.next.next;
        }
        return res.next;
    }
}

注意：循环中cur.next = cur.next.next这类的操作，就表示链表的结构已经发生改变，要从新的链表结构去找相应结点，而不是旧的结构，用下面这个代码，事先就缓存了旧结构的位置，就不用考虑当次循环下，新结构中相对位置，建议使用下面这种更易理解，上面一种容易绕晕。（画图理解）

class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode virtueNode = new ListNode(0,head);
        ListNode tem = virtueNode;
        while(tem.next!=null&&tem.next.next!=null){
            ListNode node1 = tem.next;
            ListNode node2 = tem.next.next;
            tem.next = node2;
            node1.next = node2.next;
            node2.next = node1;
            tem = node1;
        }
        return virtueNode.next;
    }
}