92 Reverse Linked List II – Medium
Problem:
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.
Thoughts:
Pointer modification. The best strategy to deal with linked list, pointer modification problem is to draw the data structure on a paper. Plus, be careful to take care of edge case. Adding a fake head is efficient to deal with edge cases most of the time.
Solutions:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if (head == null) {
return head;
}
ListNode fakeHead = new ListNode(-1);
fakeHead.next = head;
ListNode mp = fakeHead;
for (int i = 0; i < m - 1; i ++) {
mp = mp.next;
}
//mp.next is the first element to be reversed
ListNode rhead = mp;
mp = mp.next;
ListNode rtail = mp;
ListNode next = null;
for (int i = 0; i < n - m + 1; i ++) {
next = mp.next;
mp.next = rhead.next;
rhead.next = mp;
mp = next;
}
rtail.next = mp;
return fakeHead.next;
}
}