402. Remove K Digits

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible.

Note:
The length of num is less than 10002 and will be ≥ k.
The given num does not contain any leading zero.
Example 1:

Input: num = "1432219", k = 3
Output: "1219"
Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.

Example 2:

Input: num = "10200", k = 1
Output: "200"
Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.

Example 3:

Input: num = "10", k = 2
Output: "0"
Explanation: Remove all the digits from the number and it is left with nothing which is 0.

方法1: monotonic stack

水中的鱼：http://fisherlei.blogspot.com/2017/07/leetcode-remove-k-digits-solution.html

思路：

易错点：

1. 如果string本身为increasing序列，单调栈算法是舍不掉任何数的：最后要再检查一下有没有将k个数字除干净，如果没有，我们已知结果是个单调数，从后面开始pop较大值。
2. 最后要去掉前缀0。
3. 如果为空返回“0”。
4. 单调栈算法再pop的时候一定要随时查空。

class Solution {
public:
    string removeKdigits(string num, int k) {
    	// if need to remove all
        if (k == num.size()) {
            return "0";
        }
        stack<char> stk;
        
        for (char c : num) {
        	// keep popping while having better choice
        	// maintain an increasing sequence in the stack
            while (k && !stk.empty() && stk.top() > c) {
                stk.pop();
                k--;
            }
            stk.push(c);
        }
        // if haven't used up the k，delete from the end (large ones)
        // corner case like "1111"
        while (k) {
            stk.pop();
            k--;
        }
        // construct the number from the stack
        string ans;
        while (!stk.empty()) {
            ans += stk.top();
            stk.pop();
        }
        //remove all the 0 at the head
        while (ans.size() > 1 && ans.back() == '0') {
            ans.pop_back();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

grandyang: https://www.cnblogs.com/grandyang/p/5883736.html

class Solution {
public:
    string removeKdigits(string num, int k) {
        string result = "";
        int n = num.size(), keep = n - k;
        for (char c : num) {
            while (k && result.size() && result.back() > c) {
                result.pop_back();
                k--;
            }
            result.push_back(c);
        }
        result.resize(keep);
        while (!result.empty() && result[0] == '0') 
            result.erase(result.begin());
        return result.empty() ? "0" : result;
    }
};

二刷：

class Solution {
public:
    string removeKdigits(string num, int k) {
        string res = "";
        for (char c: num) {
            while (k > 0 && res.size() && res.back() > c) {
                res.pop_back();
                k--;
            }
            res.push_back(c);
        }
        
        while (k-- > 0) res.pop_back();
        
        while (!res.empty() && *res.begin() == '0') res.erase(res.begin());
        return res.empty() ? "0" : res;
    }
};