LeetCode - Reverse Linked List II
Problem description
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
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Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL
Analysis
Draw some graph may help to understand this problem.
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode p = dummy;
int i = 1;
ListNode s = dummy;
ListNode e = dummy;
while (p.next != null){
ListNode next = p.next;
if (i == m){
s = p;
e = p.next; // unused pointer
p = next;
}
else if (i > m && i <= n ){
ListNode first = s.next;
p.next = next.next;
next.next = first;
s.next = next;
}
else {
p = next;
}
i++;
}
return dummy.next;
}
}