92. Reverse Linked List II

Reverse Linked List II

代码

public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if(head == null){
return null;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
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;
}
}