Reverse Linked List II

Change-Id: I89ec151fb0967b458f61643623a5ba5de943d993

Author: applewjg

ReverseLinkedListII.java

@@ -0,0 +1,45 @@
+/*
+ Author:     King, wangjingui@outlook.com
+ Date:       Jan 2, 2015
+ Problem:    Reverse Linked List II
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/reverse-linked-list-ii/
+ Notes:
+ 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.
+
+ Solution: in-place & one-pass.
+ */
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode reverseBetween(ListNode head, int m, int n) {
+        ListNode dummy = new ListNode(-1);
+        dummy.next = head;
+        ListNode first = dummy;
+        for (int i = 0; i < m - 1; ++i) first = first.next;
+        ListNode cur = first.next;
+        for (int i = 0; i < n - m; ++i) {
+            ListNode move = cur.next;
+            cur.next = move.next;
+            move.next = first.next;
+            first.next = move;
+        }
+        return dummy.next;
+    }
+}