203.移除链表元素 leetcode.com/problems/re…
思路： 定义一个虚拟节点指向头节点，这样就能统一删除节点的操作了。
/**
* 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;
// 为了能统一删除节点的操作
ListNode dummyHead = new ListNode(-1, head);
// cur是在head前面的
ListNode cur = dummyHead;
while (cur.next != null) {
if (cur.next.val == val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
head = dummyHead.next;
return head;
}
}
707.设计链表 leetcode.com/problems/de…
class MyLinkedList {
private int size;
private ListNode head;
public MyLinkedList() {
size = 0;
head = new ListNode(0);
}
public int get(int index) {
if (size == 0) return -1;
if (index >= size) return -1;
ListNode cur = head;
while (index >= 0) {
cur = cur.next;
index--;
}
return cur.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;
size++;
ListNode pre = head;
for (int i = 0; i < index; i++) {
pre = pre.next;
}
ListNode tmp = new ListNode(val);
tmp.next = pre.next;
pre.next = tmp;
}
public void deleteAtIndex(int index) {
if (index >= size) return;
size--;
if (index == 0) {
head = head.next;
return;
}
ListNode pre = head;
for (int i = 0; i < index ; i++) {
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);
*/
206.反转链表 leetcode.com/problems/re…
递归做法 这个是典型的递归基础题了，递归就是要有黑盒的思想，不要一开始就扣细节，先有宏观思想，再到细节怎么做 Base case：就是只剩最后一个节点了，直接返回就行
/**
* 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 last = reverseList(head.next);
// 就是改变当前节点的指向嘛
// 后面的指向当前节点
head.next.next = head;
// 当前节点指向左边
head.next = null;
return last;
}
}
