206. 反转链表
206. 反转链表
1、题目介绍
反转一个单链表。
试题链接：https://leetcode-cn.com/problems/reverse-linked-list/
2、双指针做法
2.1、java
public static ListNode reverseList(ListNode head) {
if(head == null) return null;
//构造头指针
ListNode index = new ListNode(-1);
index.next = head;
ListNode pCurrent = head;
while (pCurrent.next != null) {
//使辅助移动到最后
pCurrent = pCurrent.next;
}
while (index.next != pCurrent) {
ListNode temp = index.next;
index.next = temp.next;
temp.next = pCurrent.next;
pCurrent.next = temp;
}
return index.next;
}
输出结果：
2.2、C
struct ListNode* reverseList(struct ListNode* head){
if(head == NULL) return NULL;
//构造头指针
struct ListNode* index = (struct ListNode*)malloc(sizeof(struct ListNode));
index->val = -1;
index->next = head;
struct ListNode* pCurrent = head;
while (pCurrent->next != NULL) {
//使辅助移动到最后
pCurrent = pCurrent->next;
}
while (index->next != pCurrent) {
struct ListNode* temp = index->next;
index->next = temp->next;
temp->next = pCurrent->next;
pCurrent->next = temp;
}
return index->next;
}
输出结果：
3、递归做法
3.1、java
public static ListNode reverseList(ListNode head) {
if(head == null || head.next == null) return head;
// 0、1->2->3->4->5->null
// 1、2->3->4->5->null
// 2、3->4->5->null
// 3、4->5->null
// 4、5->null
// 5
ListNode newNode = reverseList(head.next);
// System.out.println(head);
head.next.next = head;
head.next = null;
return newNode;
}
输出结果：
3.2、C
struct ListNode* reverseList(struct ListNode* head){
if(head == NULL || head->next == NULL) return head;
struct ListNode* newNode = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return newNode;
}
输出结果：
