diff --git a/README.md b/README.md
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+++ b/README.md
@@ -235,6 +235,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 312 |[Burst Balloons](https://leetcode.com/problems/burst-balloons/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/312.%20Burst%20Balloons.md)|Hard|  |
 | 313 |[Super Ugly Number](https://leetcode.com/problems/super-ugly-number/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/313.%20Super%20Ugly%20Number.md)|Medium|  |
 | 315 |[Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/)|[C++](solutions/315.%20Count%20of%20Smaller%20Numbers%20After%20Self.md)|Hard|  |
+| 316 |[Remove Duplicate Letters](https://leetcode.cn/problems/remove-duplicate-letters/)|[C++](solutions/316.%20Remove%20Duplicate%20Letters.md)|Medium|  |
 | 318 |[Maximum Product of Word Lengths](https://leetcode.com/problems/maximum-product-of-word-lengths/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/318.%20Maximum%20Product%20of%20Word%20Lengths.md)|Medium|  |
 | 319 |[Bulb Switcher](https://leetcode.com/problems/bulb-switcher/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/319.%20Bulb%20Switcher.md)|Medium|  |
 | 322 |[Coin Change](https://leetcode.com/problems/coin-change/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/322.%20Coin%20Change.md)|Medium|  |
diff --git a/solutions/316. Remove Duplicate Letters.md b/solutions/316. Remove Duplicate Letters.md
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+# [316. Remove Duplicate Letters](https://leetcode.cn/problems/remove-duplicate-letters/)
+
+# 思路
+题目有两个要求：1）每个字符只出现一次；2）字典序最小。要求1比较好办，记录下每个字符出现的次数，如果出现过就去掉即可。为了满足条件2，需要**尽可能把小字符放在前面**，所以需要找到满足 s[i]>s[i+1] 且 后面还存在字符 s[i] 的下标 i，然后去掉 s[i] 。
+
+在理解这点之后，一个直观的题解思路是：对于字符串 s，从左往右遍历，找到满足条件的 s[i]，去除它后得到新的字符串 s，然后不断进行这样的循环直到不存在重复字符。但是这种解法会创建大量的中间字符串，复杂度较高。由于在删除s[i]之后，下一个需要被删除的字符串只可能在之后，所以我们其实只需要遍历一次即可找到所有s[i]并将其删除，类似与单调栈，只是有一些限制条件（因为需要保留所有出现过的字符）。
+
+为此，我们首先需要先遍历一遍s统计每个字符的剩余出现次数。然后维护一个初始为空的最终结果字符串stk，并遍历一遍s：
+1）对于当前字符 c，如果已经在stk中，说明应该去掉c（因为有更靠前的c）；
+2）否则，如果stk末尾字符大于c且stk末尾字符在s后面还会出现，那应该去掉stk末尾字符，循环直至条件不成立，然后将c加入stk；
+
+# C++
+
+```C++
+class Solution {
+public:
+    string removeDuplicateLetters(string s) {
+        if(s.size() <= 1) return s;
+
+        vector<int>remain(256, 0);
+        for(char c: s) remain[c]++;  // 标记字符c剩余多少个
+
+        string stk;
+        vector<bool>in_stk(256, false); // 是否已经在stk中
+        for(char c: s){
+            if(!in_stk[c]){  // 如果不在stk中
+                while(!stk.empty() && stk.back() > c){  //如果c比栈顶字符小
+                    if(remain[stk.back()] == 0) break;  // 如果后面没有栈顶字符,那么不应该pop
+                    in_stk[stk.back()] = false;
+                    stk.pop_back();
+                }
+                stk.push_back(c);
+                in_stk[c] = true;
+            }
+            remain[c]--;
+        }
+        return stk;
+    }
+};
+```
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