Added Problem 23. Merge K Sorted Lists

Problem 23 Added

Author: mkhuzaima

0023_Merge_k_Sorted_Lists.java

@@ -0,0 +1,95 @@
+// id: 23
+// Name: Merge k Sorted Lists
+// link: https://leetcode.com/problems/merge-k-sorted-lists/
+// Difficulty: Hard
+
+class Solution {
+    public ListNode mergeKLists(ListNode[] lists) {
+        if (lists == null || lists.length == 0) return null;
+        return mergeListsHelper(lists, 0, lists.length - 1);
+    }
+
+    private ListNode mergeListsHelper(ListNode[] lists, int start, int end) {
+        int mid = (start + end) / 2;
+        if (start >= end) return lists[start];
+
+        ListNode first = mergeListsHelper(lists, start, mid);
+        ListNode second = mergeListsHelper(lists, mid+1, end);
+        return merge2Lists(first, second);
+    }
+
+    private ListNode merge2Lists(ListNode first, ListNode second) {
+        // make dummy node for easier addition
+        ListNode head = new ListNode(-1);
+        ListNode tail = head;
+        
+        
+        while (first != null && second != null) {
+            if (first.val < second.val) {
+                tail.next = first;
+                first = first.next;
+            }
+            else {
+                tail.next = second;
+                second = second.next;
+            }
+            tail = tail.next;
+        }
+
+        // only 1 of the following if will run
+        if (first != null) {
+            tail.next = first;
+        }
+        if (second != null) {
+            tail.next = second;
+        }
+
+        return head.next;
+    }
+  
+  
+  
+  
+    //////////////////////////////////// SECOND APPROACH
+    public ListNode mergeKLists__approach2(ListNode[] lists) {
+
+        // create a list object for merged lists
+        ListNode head = null;
+        ListNode tail = null;
+
+        while (true) {
+            // find the minimum elment from lists
+            int minNodeIndex = -1;
+
+            // minimum can also be fined using Priority Queue
+            // but PrioriryQueue will use extra space
+
+            for (int i = 0; i < lists.length; i++) {
+                if (lists[i] == null) continue;
+
+                if (
+                    minNodeIndex == -1 || // not initialized
+                    lists[i].val < lists[minNodeIndex].val 
+                ){
+                    minNodeIndex = i;
+                }
+            }
+
+            if (minNodeIndex == -1) {
+                break; // all lists are merged
+            }
+            else {
+                if ( head == null ) {
+                    head = tail = lists[minNodeIndex];
+                }
+                else {
+                    tail.next = lists[minNodeIndex];
+                    tail = tail.next;
+                }
+                lists[minNodeIndex] = lists[minNodeIndex].next;
+            }
+        }
+        
+        return head;
+    }
+}