LeetCode206反转链表、24两两交换节点
206. 反转链表
反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
solution 1:循环迭代
class Solution {
public ListNode reverseList(ListNode head) {
//循环迭代, 三指针
ListNode prev = null;
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
}
class Solution {
public ListNode reverseList(ListNode head) {
//循环迭代,三指针,head代替一指针
ListNode pre = null;
while (head != null) {
ListNode next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
solution 2:递归
class Solution {
public ListNode reverseList(ListNode head) {
//递归
if (head == null || head.next == null) return head;
ListNode p = reverseList(head.next);
head.next.next = head;
head.next = null;
return p;
}
}
24. 两两交换链表中的节点
给定一个链表,两两交换其中相邻的节点,并返回交换后的链表。
你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例:
给定 1->2->3->4, 你应该返回 2->1->4->3.
solution1 206. 反转链表
反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
solution 1:循环迭代
class Solution {
public ListNode reverseList(ListNode head) {
//循环迭代, 三指针
ListNode prev = null;
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
}
class Solution {
public ListNode reverseList(ListNode head) {
//循环迭代,三指针,head代替一指针
ListNode pre = null;
while (head != null) {
ListNode next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
solution 2:递归
class Solution {
public ListNode reverseList(ListNode head) {
//递归
if (head == null || head.next == null) return head;
ListNode p = reverseList(head.next);
head.next.next = head;
head.next = null;
return p;
}
}
24. 两两交换链表中的节点
给定一个链表,两两交换其中相邻的节点,并返回交换后的链表。
你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例:
给定 1->2->3->4, 你应该返回 2->1->4->3.
solution1 迭代iteration
//整三个指针,一个
class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode curr = head;
ListNode next = null;
ListNode tail = null;
ListNode re = head.next;
while (curr != null && curr.next != null) {
next = curr.next;
curr.next = next.next;
next.next = curr;
if (tail != null) tail.next = next;
tail = curr;
curr = curr.next;
}
return re;
}
}
solution2 递归
//ac 0ms的迭代?
class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode first = head;
ListNode second = head.next;
first.next = swapPairs(second.next);
second.next = first;
return second;
}
}
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