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.

解题思路为：

1. 顺序扫描string，当k>0时，若前一个数字比当前数字大时，将前一个数字删除，同时k--；
2. 若当前数字不是leading zero时，添加至StringBuilder；
3. 若string扫描完后，k>0，则将后k个数字删除（此时需注意删除的个数为min(k, sb.length())，因为如果存在leading zero，则sb的长度会比k小）。

public class Solution {
    public String removeKdigits(String num, int k) {
        StringBuilder sb = new StringBuilder();
        for(char c : num.toCharArray()) {
            while(k > 0 && sb.length() != 0 && sb.charAt(sb.length() - 1) > c) {
                sb.deleteCharAt(sb.length() - 1);
                k--;
            }
            if(sb.length() != 0 || c != '0') sb.append(c);
        }
        if(k > 0) sb.setLength(sb.length() - Math.min(k, sb.length()));
        return sb.length() == 0 ? "0" : sb.toString();
    }
}