diff --git a/README.md b/README.md
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@@ -293,3 +293,4 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 665 |[Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/665.%20Non-decreasing%20Array.md)|Easy|  |
 | 714 |[Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee.md)|Medium|  |
 | 905 |[Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/)|[C++](solutions/905.%20Sort%20Array%20By%20Parity.md)|Easy|  |
+| 946 |[Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/)|[C++](solutions/946.%20Validate%20Stack%20Sequences.md)|Medium|  |
diff --git a/solutions/946. Validate Stack Sequences.md b/solutions/946. Validate Stack Sequences.md
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+# [946. Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/)
+
+# 思路
+
+输入两个整数序列，第一个序列表示栈的入栈顺序，请判断第二个序列是否可能为该栈的出栈顺序。
+
+## 思路一
+
+例
+```
+pushed = [1,2,3,4,5]
+popped = [4,5,3,2,1]
+```
+出栈序列第一个是4，我们要什么时候pop出4呢，很明显是当4第一次出现在栈顶的时候（即push了4就要立马pop出），否则为了pop出4需要先pop出在4上面的元素，不符合出栈序列要求。出栈序列后面的元素同理，即是一个**贪心**的过程: 从前往后依次将元素入栈，在这个过程中不断判断栈顶元素是否为下一个popped元素，若是则应该出栈。最后如果栈为空则说明全部出栈了，返回真；否则返回假。
+
+时空复杂度均为O(n)
+
+## 思路二
+
+一开始想到的是暴力法，即模拟出入栈的过程，用DFS回溯。亲测比思路一慢很多很多，就不细说了。
+
+# C++
+
+## 思路一
+``` C++
+class Solution {
+public:
+    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
+        int n = pushed.size(), pop_i = 0;
+        stack<int>stk;
+        for(int i = 0; i < n; i++){
+            stk.push(pushed[i]);
+            while(!stk.empty() && stk.top() == popped[pop_i]){
+                stk.pop();
+                pop_i++;
+            }
+        }
+        return stk.empty();
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+private:
+    vector<int>pushed_, popped_;
+    int n;
+    bool helper(stack<int>&stk, int push_i, int pop_i){
+        if(pop_i == n) return true;
+        
+        // 出栈
+        if(!stk.empty() && stk.top() == popped_[pop_i]){ 
+            int tmp = stk.top(); stk.pop();
+            if(helper(stk, push_i, pop_i + 1)) return true;
+            stk.push(tmp);
+        }
+        
+        // 入栈
+        if(push_i >= n) return false;
+        stk.push(pushed_[push_i]);
+        if(helper(stk, push_i + 1, pop_i)) return true;
+        
+        return false;
+    }
+public:
+    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
+        pushed_ = pushed;
+        popped_ = popped;
+        n = pushed.size();
+        if(n <= 1) return true;
+        stack<int>stk, out;
+        int i = 0;
+        return helper(stk, 0, 0);
+    }
+};
+```
+