Add solution 023

Author: yanglbme

README.md

@@ -30,6 +30,7 @@ Complete solutions to Leetcode problems, updated daily.
 
 | # | Title | Tags |
 |---|---|---|
+| 023 | [Merge k Sorted Lists](https://github.com/yanglbme/leetcode/tree/master/solution/023.Merge%20k%20Sorted%20Lists) | `Linked List`, `Divide and Conquer`, `Heap` |
 
 ## Contributions
 I'm looking for long-term contributors/partners to this repo! Send me PRs if you're interested! See the following:

solution/023.Merge k Sorted Lists/README.md

@@ -0,0 +1,69 @@
+## 合并K个排序链表
+### 题目描述
+
+合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。
+
+示例:
+```
+输入:
+[
+  1->4->5,
+  1->3->4,
+  2->6
+]
+输出: 1->1->2->3->4->4->5->6
+```
+
+### 解法
+从链表数组索引 0 开始，[合并前后相邻两个有序链表](https://github.com/yanglbme/leetcode/tree/master/solution/021.Merge%20Two%20Sorted%20Lists)，放在后一个链表位置上，依次循环下去...最后 lists[len - 1] 即为合并后的链表。注意处理链表数组元素小于 2 的情况。
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode mergeKLists(ListNode[] lists) {
+        if (lists == null || lists.length == 0) {
+            return null;
+        }
+        
+        int len = lists.length;
+        if (len == 1) {
+            return lists[0];
+        }
+       
+        // 合并前后两个链表，结果放在后一个链表位置上，依次循环下去
+        for (int i = 0; i < len - 1; ++i) {
+            lists[i + 1] = mergeTwoLists(lists[i], lists[i + 1]);
+        }
+        return lists[len - 1];
+        
+    }
+    
+    /**
+     * 合并两个有序链表
+     * @param l1 
+     * @param l2
+     * @return listNode
+     */
+    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if (l1 == null) {
+            return l2;
+        }
+        if (l2 == null) {
+            return l1;
+        }
+        if (l1.val < l2.val) {
+            l1.next = mergeTwoLists(l1.next, l2);
+            return l1;
+        }
+        l2.next = mergeTwoLists(l1, l2.next);
+        return l2;
+    }
+}
+```

solution/023.Merge k Sorted Lists/Solution.java

@@ -0,0 +1,40 @@
+class Solution {
+    public ListNode mergeKLists(ListNode[] lists) {
+        if (lists == null || lists.length == 0) {
+            return null;
+        }
+        
+        int len = lists.length;
+        if (len == 1) {
+            return lists[0];
+        }
+       
+        // 合并前后两个链表，结果放在后一个链表位置上，依次循环下去
+        for (int i = 0; i < len - 1; ++i) {
+            lists[i + 1] = mergeTwoLists(lists[i], lists[i + 1]);
+        }
+        return lists[len - 1];
+        
+    }
+    
+    /**
+     * 合并两个有序链表
+     * @param l1 
+     * @param l2
+     * @return listNode
+     */
+    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if (l1 == null) {
+            return l2;
+        }
+        if (l2 == null) {
+            return l1;
+        }
+        if (l1.val < l2.val) {
+            l1.next = mergeTwoLists(l1.next, l2);
+            return l1;
+        }
+        l2.next = mergeTwoLists(l1, l2.next);
+        return l2;
+    }
+}