add 402

README.md

@@ -287,6 +287,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 397 |[Integer Replacement](https://leetcode.com/problems/integer-replacement/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/397.%20Integer%20Replacement.md)|Medium|  |
 | 398 |[Random Pick Index](https://leetcode.com/problems/random-pick-index/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/398.%20Random%20Pick%20Index.md)|Medium|  |
 | 400 |[Nth Digit](https://leetcode.com/problems/nth-digit)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/400.%20Nth%20Digit.md)|Medium|  |
+| 402 |[Remove K Digits](https://leetcode.cn/problems/remove-k-digits/)|[C++](solutions/402.%20Remove%20K%20Digits.md)|Medium|  |
 | 404 |[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/404.%20Sum%20of%20Left%20Leaves.md)|Easy|  |
 | 405 |[Convert a Number to Hexadecimal](https://leetcode.com/problems/convert-a-number-to-hexadecimal)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/405.%20Convert%20a%20Number%20to%20Hexadecimal.md)|Easy|  |
 | 406 |[Queue Reconstruction by Height](https://leetcode.com/problems/queue-reconstruction-by-height/)|[C++](solutions/406.%20Queue%20Reconstruction%20by%20Height.md)|Medium|  |

solutions/402. Remove K Digits.md

@@ -0,0 +1,38 @@
+# [402. Remove K Digits](https://leetcode.cn/problems/remove-k-digits/)
+
+# 思路
+要求返回的数字最小，其实就是字典序最小，所以类似[316. Remove Duplicate Letters](https://leetcode.cn/problems/remove-duplicate-letters/)：需要**尽可能把小数字放在前面**，所以需要找到满足 num[i]>num[i+1]，然后去掉 num[i]，即单调栈。
+
+具体的，维护一个初始为空的最终结果字符串stk，并遍历一遍num：
+1）对于当前字符 c，如果stk结尾字符比c大，说明找到了 num[i]>num[i+1]，即应该去掉skt结尾字符，并k--，然后继续判断直到不满足；
+2）将c加入stk；
+
+注意最后需要处理k还大于0以及前导0。
+
+关键词：单调栈
+
+# C++
+```C++
+class Solution {
+public:
+    string removeKdigits(string num, int k) {
+        if(num.size() <= 1) return "0";
+        string stk = "";
+        for(char c: num){
+            while(!stk.empty() && k > 0 && c < stk.back()){
+                k--;
+                stk.pop_back();
+            }
+            stk.push_back(c);
+        }
+
+        for(int i = 0; i < k; i++) stk.pop_back();
+        int lead_0_cnt = 0;
+        for(char c: stk){
+            if(c != '0') break;
+            else lead_0_cnt++;
+        }
+        return lead_0_cnt == stk.size() ? "0" : stk.substr(lead_0_cnt);
+    }
+};
+```