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diff --git a/‎115. Merge k Sorted Lists/main.cpp
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+这一阶段的 LeetCode 有一个明显的特点，就是在增加难度的同时，还紧密联系之前的 Easy 难度的题目。
+
+昨天的两道姊妹题是如此，今天的链表题依旧如此。
+
+这道题，显然是 [019. Merge Two Sorted Lists](../019. Merge Two Sorted Lists) 的升级版。
+
+对于想复习基础的链表操作的童鞋，我推荐看看我的[这篇总结](http://segmentfault.com/blog/pezy/1190000002490878)。
+
+-----
+
+拿到问题，合并多个链表。而此刻，我们已经掌握了合并两个链表的方法。于是我们很自然的列出以下几种情况：
+
+- lists.empty() : 返回 NULL
+- lists.size() == 1 : 返回该链表
+- lists.size() == 2 : 正好碰枪口，mergeTwoLists.
+- lists.size() == 3 : 先合并后两个，在将其结果与第三个合并。
+- lists.size() == 4 : 先两两合并，然后再将结果合并。
+- ...
+
+就没必要一直列下去了吧。
+
+分而治之，是非常自然而然的思路，我们如果会造枪，但要的却是炮，我们将枪捆在一起，捆的无限多，便成了炮。
+
+注意，为了更方便分治，我们将原题目的接口扩展为更通用的迭代器接口。
+
+然后核心的语句是：
+
+    return mergeTwoLists(mergeKLists(beg, beg + std::distance(beg, end)/2), mergeKLists(beg + std::distance(beg, end)/2, end));
+
+提交，AC，离最快差了不过 4 ms.
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+#include "solution.h"
+#include <iostream>
+
+using namespace std;
+
+ListNode *create_linkedlist(std::initializer_list<int> lst)
+{
+    auto iter = lst.begin();
+    ListNode *head = lst.size() ? new ListNode(*iter++) : nullptr;
+    for (ListNode *cur=head; iter != lst.end(); cur=cur->next)
+        cur->next = new ListNode(*iter++);
+    return head;
+}
+
+void clear(ListNode *head)
+{
+    while (head) {
+        ListNode *del = head;
+        head = head->next;
+        delete del;
+    }
+}
+
+void printList(ListNode *head) {
+    for (ListNode *cur = head; cur; cur = cur->next)
+        cout << cur->val << "->";
+    cout << "\b\b  " << endl;
+}
+
+int main()
+{
+    Solution s;
+    vector<ListNode *> vec{
+        create_linkedlist({1,3,5,7,9}),
+        create_linkedlist({2,4,6,8,10}),
+        create_linkedlist({0,11,12,13,14})
+    };
+    ListNode *ret = s.mergeKLists(vec);
+    printList(ret);
+    clear(ret);
+
+    return 0;
+}
+
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+#include <vector>
+using std::vector;
+#include <cstddef>
+
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+};
+
+class Solution {
+    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
+        ListNode *head = NULL, **lastPtrRef = &head;
+        for (;l1 && l2; lastPtrRef = &((*lastPtrRef)->next)) {
+            if (l1->val <= l2->val) { *lastPtrRef = l1; l1 = l1->next; }
+            else { *lastPtrRef = l2; l2 = l2->next; }
+        }
+        *lastPtrRef = l1 ? l1 : l2;
+        return head;
+    }
+
+    using vecNodeCIter = vector<ListNode *>::const_iterator;
+    ListNode *mergeKLists(vecNodeCIter beg, vecNodeCIter end) {
+        if (beg == end) return NULL;
+        else if (std::distance(beg, end) == 1) return *beg;
+        else if (std::distance(beg, end) == 2) return mergeTwoLists(*beg, *std::next(beg));
+        else return mergeTwoLists(mergeKLists(beg, beg + std::distance(beg, end)/2), mergeKLists(beg + std::distance(beg, end)/2, end));
+    }
+public:
+    ListNode *mergeKLists(vector<ListNode *> &lists) {
+        return mergeKLists(lists.cbegin(), lists.cend());
+    }
+};