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.

根本还是单链表的翻转，只不过需要预先保存好一些节点的值

public class ReverseLinkedListII {
public ListNode reverseBetween(ListNode head, int m, int n) {

if(head==null || m>n) {
return null;
}

//设置一个dummyNode 便于操作
ListNode dummyNode = new ListNode(-1);
dummyNode.next = head;
head = dummyNode;
//到m-n区间前面一个节点
for(int i=1;i<m;i++) {
if(head==null) {
return null;
}
head = head.next;
}

//保存n-m区间前一个节点
ListNode pre = head;
//保存n-m区间第一个节点
ListNode fNode = head.next;
ListNode lastNode = null;
ListNode curNode= fNode;
for(int i=m;i<=n;i++) {
ListNode tmp = curNode.next;
curNode.next = lastNode;
lastNode = curNode;
curNode = tmp;
}
pre.next = lastNode;
fNode.next = curNode;

return dummyNode.next;

}
}