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.

solutions:

本題意是要求給定一個表示為字串的非負整數num，從該數字中刪除k個數字，以便新的數字是最小的。
 

我們能想到的是給定的字串不能改變相對位置，只能刪除其中的數字。所以很容易知道可以從最高位開始。比如6位num， k = 2.那麼刪除後剩下4位，那麼最高位肯定是在前三位中取（尤其地若在第三位取最高位，則字串就確定了）

string removeKdigits(string num, int k) {
    string res;
    helper(num, k, res);
    return res;
}
void helper(string num, int k, string& res) {
    if (k == num.size()) {
        if (res.size() == 0) res = "0";
        return;
    }
    if (k == 0) {
        res += num;
        if (res.size() == 0) res = "0";
        return;
    }
    int minid = 0;
    for (int i = 0; i <= k; i++) {
        if (num[i] < num[minid]) minid = i;
    }
    if (res.size() != 0 || num[minid] != '0') res += num[minid];
    helper(num.substr(minid+1, num.size()-minid-1), k-minid, res);
    return;
}