Remove K Digits （第六周 贪心算法）

问题描述

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.

Example1:

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.

Example2:

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.

Example3:

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

算法思路

主要是使用了贪心算法。
（1）基本思想就是用栈去维护一个递增的序列，依次遍历全部字符（数字），如果当前的数字比栈顶的元素小，就把栈顶的元素删除，直到栈顶的元素比当前数字大或者为空，然后就可以把当前的数字压入栈顶。
（2）最后我们就可以得到一个小的序列，然后我们只需要取出栈中前len – k个数字就可以了。
（3）要注意的是，得到前len – k个数字后，要去除掉前面的0元素。并且当得到的栈为空的时候，手动输出0。

算法代码

class Solution {
public:
    string removeKdigits(string num, int k) {
        string res = "";
        int n = num.size(), keep = n - k;

        for (int i = 0; i < n; i++) {
            while (k && res.size() && res.back() > num[i]) {
                res.pop_back();
                --k;
            }
            res.push_back(num[i]);
        }
        res.resize(keep);
        while (!res.empty() && res[0] == '0') res.erase(res.begin());
        return res.empty() ? "0" : res;
    }
};