diff --git a/README.md b/README.md
index a6db914..1d42278 100644
--- a/README.md
+++ b/README.md
@@ -178,6 +178,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 215 |[Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/215.%20Kth%20Largest%20Element%20in%20an%20Array.md)|Medium|  |
 | 216 |[Combination Sum III](https://leetcode.com/problems/combination-sum-iii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/216.%20Combination%20Sum%20III.md)|Medium|  |
 | 217 |[Contains Duplicate](https://leetcode.com/problems/contains-duplicate)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/217.%20Contains%20Duplicate.md)|Easy|  |
+| 218 |[The Skyline Problem](https://leetcode.com/problems/the-skyline-problem/)|[C++](solutions/218.%20The%20Skyline%20Problem.md)|Hard|  |
 | 219 |[Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/219.%20Contains%20Duplicate%20II.md)|Easy|  |
 | 220 |[Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/220.%20Contains%20Duplicate%20III.md)|Medium|  |
 | 221 |[Maximal Square](https://leetcode.com/problems/maximal-square/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/221.%20Maximal%20Square.md)|Medium|  |
diff --git a/solutions/218. The Skyline Problem.md b/solutions/218. The Skyline Problem.md
new file mode 100644
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@@ -0,0 +1,48 @@
+# [218. The Skyline Problem](https://leetcode.com/problems/the-skyline-problem/)
+
+# 思路
+
+扫描法
+
+想象一下从左到右扫描，如果当前扫描线穿过的building的最大高度与上一次的最大高度不一样，说明遇到了转折点，记录这个转折点即可。
+
+首先为了从左到右扫描，我们需要将所有边界（包括左边界和右边界）与高度的pair按照水平位置排序（为了区分左右边界，将左边界的高度存为负数，建立左边界和负高度的pair）；
+
+然后，我们还需要用一个容器记录当前扫描线穿过的building的所有高度，而且能够快速获得最大高度以及删除某个高度，那么`multiset`就很不错。我们先插入高度0为初始高度，然后开始扫描，如果遇到高度为负值的pair，说明是左边界，那么将高度插入multiset中；否则说明遇到了右边界，应该从multiset中删除这个高度。然后取出此时multiset中最高的高度，判断是否等于上一轮的最大高度，若不是则表示遇到了转折，需要存放到结果数组中。
+
+扫描过程动图可参考[此处](https://leetcode-cn.com/problems/the-skyline-problem/solution/218tian-ji-xian-wen-ti-sao-miao-xian-fa-by-ivan_al/)
+
+时间复杂度O(nlogn)；空间复杂度O(n)
+
+# C++
+``` C++
+class Solution {
+public:
+    vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
+        vector<pair<int,int>>all;
+        vector<vector<int>>res;
+        multiset<int>st;
+        
+        for(auto &b: buildings){
+            all.push_back({b[0], -b[2]}); // 左边界, -高度
+            all.push_back({b[1], b[2]});  // 右边界, 高度
+        }
+        sort(all.begin(), all.end()); // 将边界排好序
+        
+        int pre_h = 0, cur_h;
+        st.insert(0);
+        for(auto &b: all){ // 从左至右扫描
+            if(b.second < 0) st.insert(-b.second);
+            else st.erase(st.find(b.second));
+            
+            cur_h = *st.rbegin(); // 当前最高高度
+            if(cur_h != pre_h){
+                res.push_back({b.first, cur_h});
+                pre_h = cur_h;
+            }
+        }
+        return res;
+    }
+};
+```
+