leetcode92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:

Given m, n satisfy the following condition:

1 ≤ m ≤ n ≤ length of list.

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思路：将中间部分的链表逆序即可。

总结: 对于链表的操作，可以加个头结点使得操作更加的方便。

我的方案：
public static ListNode solution(ListNode head, int m, int n) {

if (n == m)
return head;

ListNode prev, current;
prev = current =head;
if (m == 1) {
// 为链表设置个头结点， 更容易操作，主要是这里需要对m==1时 prev进行定义。 我们可以就加头结点,不区分m是否=1
prev = new ListNode(0);
prev.next = head;
head = prev;
}

int i =1;
for (;i<m;i++)
{
prev = current;
current = current.next;
}
ListNode tmp = current;

while (i++!=n)
{
current = tmp.next;
tmp.next = current.next;
current.next = prev.next;
prev.next = current;

//对中间部分翻转，注意指针操作的顺序，画图可以更清晰
}

if (m == 1)
return head.next;
return head;

}

public ListNode reverseBetween(ListNode head, int m, int n) {
if(head == null) return null;
ListNode dummy = new ListNode(0);
// create a dummy node to mark the head of this list
dummy.next = head;
// make a pointer pre as a marker for the node before reversing
for(int i = 0; i<m-1; i++) pre = pre.next;

ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
ListNode then = start.next; // a pointer to a node that will be reversed

// 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
// dummy-> 1 -> 2 -> 3 -> 4 -> 5

for(int i=0; i<n-m; i++)
{
start.next = then.next;
then.next = pre.next;
pre.next = then;
then = start.next;
}

// first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
// second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)

return dummy.next;

}