diff --git a/README.md b/README.md
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+++ b/README.md
@@ -30,6 +30,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
| 20 |[Valid Parentheses](https://leetcode.com/problems/valid-parentheses)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/20.%20Valid%20Parentheses.md)|Easy| |
| 21 |[Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/21.%20Merge%20Two%20Sorted%20Lists.md)|Easy| |
| 22 |[Generate Parentheses](https://leetcode.com/problems/generate-parentheses)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/22.%20Generate%20Parentheses.md)|Medium| |
+| 23 |[Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)|[C++](solutions/23.%20Merge%20k%20Sorted%20Lists.md)|Hard| |
| 24 |[Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/24.%20Swap%20Nodes%20in%20Pairs.md)|Medium| |
| 26 |[Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/26.%20Remove%20Duplicates%20from%20Sorted%20Array.md)|Easy| |
| 27 |[Remove Element](https://leetcode.com/problems/remove-element)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/27.%20Remove%20Element.md)|Easy| |
diff --git a/solutions/23. Merge k Sorted Lists.md b/solutions/23. Merge k Sorted Lists.md
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+# [23. Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)
+
+# 思路
+
+将k个排好序的链表合并成一个大的有序链表。
+
+## 思路一
+
+最好想的就是维护一个堆(priority_queue)或者红黑树(multiset或者multimap)的数据结构。首先把k个链表的首元素都加入其中,然后每次取出最小的那个元素加入最终结果的链表中,然后把取出元素所在链表的下一个元素再加入结构中,以此类推,直到结构中没有元素了。
+
+
+设所有元素个数为N,时间复杂度O(Nlogk),空间复杂度O(k)
+
+## 思路二
+
+另一个思路就是外排序,即k路归并排序的思想,如下图所示
+
+
+

+
+
+由于每次归并时间复杂度是线性的,所以每一轮归并时间均为O(N),而需要 logk 轮归并,所以时间复杂度O(Nlogk)。
+
+空间复杂度O(1)
+
+# C++
+## 思路一、使用priority_queue
+**注意学习如何自定义compare类**
+``` C++
+// 默认为最大堆, 所以需要自定义compare类使成为最小堆
+struct compare {
+ bool operator()(const ListNode* p1, const ListNode* p2) {
+ return p1 -> val > p2 -> val;
+ }
+};
+
+class Solution {
+public:
+ ListNode* mergeKLists(vector& lists) {
+ priority_queue, compare>min_heap;
+
+ for(int i = 0; i < lists.size(); i++)
+ if(lists[i]) min_heap.push(lists[i]);
+
+ if(min_heap.empty()) return NULL;
+
+ ListNode *head = min_heap.top(), *p = head;
+ min_heap.pop();
+ if(p -> next) min_heap.push(p -> next);
+ while(!min_heap.empty()){
+ p -> next = min_heap.top(); min_heap.pop();
+ p = p -> next;
+ if(p -> next) min_heap.push(p -> next);
+ }
+ return head;
+ }
+};
+```
+
+## 思路一、使用multimap
+**注意multimap插入时需传入一个{key, value}的pair**
+``` C++
+class Solution {
+public:
+ ListNode* mergeKLists(vector& lists) {
+ multimapmp;
+ for(int i = 0; i < lists.size(); i++)
+ if(lists[i]) mp.insert({lists[i] -> val, lists[i]});
+
+ if(mp.empty()) return NULL;
+
+ ListNode *head = mp.begin() -> second, *p = head;
+ mp.erase(mp.begin());
+ if(p -> next) mp.insert({p -> next -> val, p -> next});
+ while(!mp.empty()){
+ p -> next = mp.begin() -> second;
+ mp.erase(mp.begin());
+ p = p -> next;
+ if(p -> next) mp.insert({p -> next-> val, p -> next});
+ }
+ return head;
+ }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+private:
+ ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+ /*copy from 21. Merge Two Sorted Lists */
+ ListNode *res = new ListNode(0);
+ ListNode *pre = res;
+ while(l1 && l2){
+ if(l1 -> val <= l2 -> val){
+ pre -> next = l1;
+ l1 = l1 -> next;
+ }
+ else{
+ pre -> next = l2;
+ l2 = l2 -> next;
+ }
+ pre = pre -> next;
+ }
+
+ if(l1) pre -> next = l1;
+ else pre -> next = l2;
+
+ ListNode *head = res;
+ res = head -> next;
+ head -> next = NULL;
+ delete head;
+ return res;
+ }
+
+public:
+ ListNode* mergeKLists(vector& lists) {
+ int n = lists.size();
+ if(!n) return NULL;
+
+ int interval = 1;
+ while(interval < n){
+ for(int i = 0; i < n - interval; i += (2 * interval))
+ lists[i] = mergeTwoLists(lists[i], lists[i+interval]);
+
+ interval*=2;
+ }
+ return lists[0];
+ }
+};
+```
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