refactor 92

fishercoder1534 · fishercoder1534 · commit 19f54ff71327 · 2020-05-09T10:10:18.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_92.java b/src/main/java/com/fishercoder/solutions/_92.java
@@ -2,48 +2,39 @@
 
 import com.fishercoder.common.classes.ListNode;
 
-/**
- * 92. 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.
- */
 public class _92 {
 
-  public static class Solution1 {
-    /** credit: https://discuss.leetcode.com/topic/8976/simple-java-solution-with-clear-explanation */
-    public ListNode reverseBetween(ListNode head, int m, int n) {
-      // use four nodes, pre, start, then, dummy
-      // just reverse the nodes along the way
-      ListNode dummy = new ListNode(-1);
-      dummy.next = head;
-      ListNode pre = dummy;
-      for (int i = 0; i < m - 1; i++) {
-        pre = pre.next;
-      }
+    public static class Solution1 {
+        /**
+         * credit: https://discuss.leetcode.com/topic/8976/simple-java-solution-with-clear-explanation
+         */
+        public ListNode reverseBetween(ListNode head, int m, int n) {
+            // use four nodes, pre, start, then, dummy
+            // just reverse the nodes along the way
+            ListNode dummy = new ListNode(-1);
+            dummy.next = head;
+            ListNode pre = dummy;
+            for (int i = 0; i < m - 1; i++) {
+                pre = pre.next;
+            }
 
-      ListNode start = pre.next;// start is the node prior to reversing, in the given example,
-      // start is node with value 1
-      ListNode then = start.next;// then is the node that we'll start to reverse, in the given
-      // example, it's 2
+            ListNode start = pre.next;// start is the node prior to reversing, in the given example,
+            // start is node with value 1
+            ListNode then = start.next;// then is the node that we'll start to reverse, in the given
+            // example, it's 2
 
-      for (int i = 0; i < n - m; i++) {
-        // pay special attention to this for loop, it's assigning then.next to start.next, it
-        // didn't initialize a new node
-        // this does exactly what I desired to do, but I just didn't figure out how to implement
-        // it, thumbs up to the OP!
-        start.next = then.next;
-        then.next = pre.next;
-        pre.next = then;
-        then = start.next;
-      }
+            for (int i = 0; i < n - m; i++) {
+                // pay special attention to this for loop, it's assigning then.next to start.next, it
+                // didn't initialize a new node
+                // this does exactly what I desired to do, but I just didn't figure out how to implement
+                // it, thumbs up to the OP!
+                start.next = then.next;
+                then.next = pre.next;
+                pre.next = then;
+                then = start.next;
+            }
 
-      return dummy.next;
+            return dummy.next;
+        }
     }
-  }
 }
