reverse sub list in a linked list

codejayant · codejayant · commit 5bd802f22c82 · 2020-04-05T17:19:08.000-07:00
diff --git a/src/main/java/com/codejayant/linkedList/inPlaceReversal/ReverseSubList.java b/src/main/java/com/codejayant/linkedList/inPlaceReversal/ReverseSubList.java
@@ -0,0 +1,88 @@
+package com.codejayant.linkedList.inPlaceReversal;
+
+import com.codejayant.utils.ListNode;
+
+/**
+ * Reverse a linked list from position m to n. Do it in one-pass.
+ * <p>
+ * Note: 1 ≤ m ≤ n ≤ length of list.
+ * <p>
+ * Example:
+ * <p>
+ * Input: 1->2->3->4->5->NULL, m = 2, n = 4
+ * Output: 1->4->3->2->5->NULL
+ *
+ * @see <a href="https://leetcode.com/problems/reverse-linked-list-ii/">LeetCode Problem</a>
+ * @see <a href="https://www.educative.io/courses/grokking-the-coding-interview/qVANqMonoB2">Educative Problem</a>
+ * @see <a href="https://leetcode.com/articles/reverse-linked-list-ii/">Article</a>
+ */
+public class ReverseSubList {
+
+    /**
+     * in-place iterative reversal of sub-list
+     * T: O(n)
+     * S: O(1)
+     *
+     * @param head root node
+     * @param p    start position of list to reverse
+     * @param q    end position of list to reverse
+     * @return head of modified list
+     */
+    private static ListNode reverse(ListNode head, int p, int q) {
+        if (p == q) {
+            return head;
+        }
+
+        // after skipping 'p-1' nodes, current will point to 'p'th node
+        ListNode current = head, previous = null;
+        for (int i = 0; current != null && i < p - 1; ++i) {
+            previous = current;
+            current = current.next;
+        }
+
+        // we are interested in three parts of the LinkedList, part before index 'p', part between 'p' and
+        // 'q', and the part after index 'q'
+        ListNode lastNodeOfFirstPart = previous; // points to the node at index 'p-1'
+        // after reversing the LinkedList 'current' will become the last node of the sub-list
+        ListNode lastNodeOfSubList = current;
+        ListNode next; // will be used to temporarily store the next node
+        // reverse nodes between 'p' and 'q'
+        for (int i = 0; current != null && i < q - p + 1; i++) {
+            next = current.next;
+            current.next = previous;
+            previous = current;
+            current = next;
+        }
+
+        // connect with the first part
+        if (lastNodeOfFirstPart != null) {
+            lastNodeOfFirstPart.next = previous; // 'previous' is now the first node of the sub-list
+        } else {   // this means p == 1 i.e., we are changing the first node (head) of the LinkedList
+            head = previous;
+        }
+
+        // connect with the last part
+        if (lastNodeOfSubList != null) {
+            lastNodeOfSubList.next = current;
+        }
+
+        return head;
+    }
+
+    public static void main(String[] args) {
+        ListNode head = new ListNode(1);
+        head.next = new ListNode(2);
+        head.next.next = new ListNode(3);
+        head.next.next.next = new ListNode(4);
+        head.next.next.next.next = new ListNode(5);
+
+        ListNode result = ReverseSubList.reverse(head, 2, 4);
+        // expected value: 1 4 3 2 5
+        System.out.print("Nodes of the reversed LinkedList are: ");
+        while (result != null) {
+            System.out.print(result.val + " ");
+            result = result.next;
+        }
+    }
+
+}
