You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
124 lines
3.1 KiB
124 lines
3.1 KiB
---
|
|
链表相关
|
|
---
|
|
|
|
[06. 从头到尾打印链表](https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/)
|
|
|
|
```java
|
|
class Solution {
|
|
public int[s] reversePrint(ListNode head) {
|
|
ListNode temp = head;
|
|
int size = 0;
|
|
while (temp != null) {
|
|
temp = temp.next;
|
|
size++;
|
|
}
|
|
int[] result = new int[size];
|
|
int index = size - 1;
|
|
while (head != null) {
|
|
result[index--] = head.val;
|
|
head = head.next;
|
|
}
|
|
return result;
|
|
}
|
|
}
|
|
```
|
|
|
|
```java
|
|
class Solution {
|
|
public int[] reversePrint(ListNode head) {
|
|
Stack<ListNode> stack = new Stack<ListNode>();
|
|
ListNode temp = head;
|
|
while (temp != null) {
|
|
stack.push(temp);
|
|
temp = temp.next;
|
|
}
|
|
int size = stack.size();
|
|
int[] print = new int[size];
|
|
for (int i = 0; i < size; i++) {
|
|
print[i] = stack.pop().val;
|
|
}
|
|
return print;
|
|
}
|
|
}
|
|
```
|
|
|
|
[22. 链表中倒数第 k 个节点](https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/)
|
|
|
|
```java
|
|
class Solution {
|
|
public ListNode getKthFromEnd(ListNode head, int k) {
|
|
int length = 0;
|
|
ListNode temp = head;
|
|
while (temp != null) {
|
|
temp = temp.next;
|
|
length++;
|
|
}
|
|
for (int i = 0; i < length - k; i++) {
|
|
head = head.next;
|
|
}
|
|
return head;
|
|
}
|
|
}
|
|
```
|
|
|
|
```java
|
|
class Solution {
|
|
public ListNode getKthFromEnd(ListNode head, int k) {
|
|
ListNode h0 = head;
|
|
for (int i = 0; i < k; i++) {
|
|
h0 = h0.next;
|
|
}
|
|
while (h0 != null) {
|
|
h0 = h0.next;
|
|
head = head.next;
|
|
}
|
|
return head;
|
|
}
|
|
}
|
|
```
|
|
|
|
[24. 反转链表](https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/)
|
|
|
|
```java
|
|
class Solution {
|
|
public ListNode reverseList(ListNode head) {
|
|
if (head == null || head.next == null) {
|
|
return head;
|
|
}
|
|
ListNode h1 = head;
|
|
ListNode h2 = head.next;
|
|
ListNode h3 = null;
|
|
h1.next = null;
|
|
while (h2 != null) {
|
|
h3 = h2.next;
|
|
h2.next = h1;
|
|
h1 = h2;
|
|
h2 = h3;
|
|
}
|
|
return h1;
|
|
}
|
|
}
|
|
```
|
|
|
|
```java
|
|
class Solution {
|
|
public ListNode reverseList(ListNode head) {
|
|
// 递归终止条件是当前为空,或者下一个节点为空
|
|
if (head == null || head.next == null) {
|
|
return head;
|
|
}
|
|
// 这里的 h1 就是最后一个节点
|
|
ListNode h1 = reverseList(head.next);
|
|
// 如果链表是 1->2->3->4->5,那么此时的 cur 就是 5
|
|
// 而 head 是4,head的 下一个是 5,下下一个是空
|
|
// 所以 head.next.next 就是 5->4
|
|
head.next.next = head;
|
|
// 防止链表循环,需要将 head.next 设置为空
|
|
head.next = null;
|
|
// 每层递归函数都返回 h1,也就是最后一个节点
|
|
return h1;
|
|
}
|
|
}
|
|
```
|
|
|
|
|