leetcode 402. Remove K Digits 贪心算法 + DFS深度优先遍历

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.

题意很简单，就是去掉k个bit，然后求剩下的值的所有可能中的最小值，第一个想到的方法就是和上一道题的DFS深度优先遍历的做法，但是会超时，我隐隐约约感觉到可能使用stack等来解决，但是不会做，于是网上找了个做法，

建议和leetcode 738. Monotone Increasing Digits 最大单调递增数字 一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <bitset>using namespace std;class Solution {public:    string removeKdigits(string num, int k)    {        if (k >= num.size())            return "0";        string res = "";        int count = k;        for (char c : num)        {            while (count > 0 && res.size() > 0 && res.back() > c)            {                res.pop_back();                count--;            }            res.push_back(c);        }        res = res.substr(0,num.length()-k);        while (res.empty()== false && res[0] == '0')             res.erase(res.begin());        return res.length()<=0 ? "0" : res;    }    //和上一道题一样，就是一个DFS做法，但是会超时    vector<int> all;    string removeKdigitsByDFS(string num, int k)     {        if (k >= num.length())            return "0";        vector<int> bit(num.length(), 0);        vector<int> flag(num.length(), 0);        for (int i = 0; i < bit.size(); i++)            bit[i] = (int)(num[i] - '0');        getAll(0,k, flag, bit);        int minRes = numeric_limits<int>::max();        for (int one : all)            minRes = min(minRes, one);        return to_string(minRes);    }    void getAll(int begin,int k,vector<int>& flag,vector<int>& bit)    {        if (k == 0)        {            int a = 0;            for (int i = 0; i < bit.size(); i++)            {                if (flag[i] == 0)                    a = a * 10 + bit[i];            }            all.push_back(a);        }        else if (begin >= flag.size())            return;        else        {            getAll(begin + 1, k, flag, bit);            flag[begin] = 1;            getAll(begin + 1, k - 1, flag, bit);            flag[begin] = 0;        }    }};