402. Remove K Digits 去掉K位数字

题目链接
tag:

• Medium；
• Greedy；

question：
  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.

思路：
  这道题让我们从给定的数字去掉k位，要使得留下来的数字最小；其基本思想是利用stack维护一个递增的序列，也就是说将字符串中的字符依次入栈，如果当前字符串比栈顶元素小，并且还可以继续删除元素，那么就将栈顶元素去掉，这样就可以保证将当前元素加进去一定可以得到一个较小的序列，算做一个Greedy的思想，最后我们只取前len-k个元素构成一个序列即可，如果这样的到的是一个空串那就手动返回0，当然需要注意字符串首字符不为0。AC代码如下：

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