forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_206.java
59 lines (53 loc) · 2.12 KB
/
_206.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
package com.fishercoder.solutions;
import com.fishercoder.common.classes.ListNode;
/**
* 206. Reverse Linked List
*
* Reverse a singly linked list.*/
public class _206 {
public static class Solution1 {
/**
* creating a newHead = null is a very common/smart way to handle such cases, the logic flows out very naturally:
* create a new node called "next" to hold current head's next node
* then we could redirect head's next pointer to point to newHead which is head's previous node
* the above two steps finished the reversion, to continue this process until we reach the end of the original list,
* we'll assign current "head" to new "newHead", and current "next" to be new "head" for the next iteration, here's the code
*/
public ListNode reverseList(ListNode head) {
/**It works out the best to set up a debug point and visualize this process:
* e.g. 1->2->3-null
* at the end of the first iteration of the while loop, the status is like this:
* newHead: 1->null
* head: 2->3-null
* then it continues the iteration.*/
ListNode newHead = null;
while (head != null) {
ListNode next = head.next;
head.next = newHead;
newHead = head;
head = next;
}
return newHead;
}
}
public static class Solution2 {
/**
* following the above iterative version, the recursive solution flows out so naturally:
* basically, we just replaced the while loop with a recursive function
* still, a null newHead proves to be very helpful.
*/
public ListNode reverseList(ListNode head) {
return reverse(head, null);
}
ListNode reverse(ListNode head, ListNode newHead) {
if (head == null) {
return newHead;
}
ListNode next = head.next;
head.next = newHead;
newHead = head;
head = next;
return reverse(head, newHead);
}
}
}