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.

思路：
有一个N为长的数字，用字符串来代替了，现在要求你将它删除K位，使得其得到的结果最小。
首先这是一个贪心问题，即我们可以将问题转化为，一个长度为N的数字里面，删除哪个数可以使得数变得最小，那么如何删除呢？
1、因为数字开头不允许是0，当第二位是0的情况下下，这时候我们删除了第一位数，那么至少可以使数字小两个量级，而其他位置最多也就是小一个数量级，所以这种情况毫无疑问删除第一个，后面打头的0也自动消解 。
2、否则，我们从头开始找，找到第一个下降的数，如1234553，那么最后一个3前面的5就是，删除它得到的数字是最小的。

class Solution {
    public String removeKdigits(String num, int k) {
        int n;
        while(true){
            n = num.length();
            if(n <= k || n == 0) 
                return "0";
            if(k-- == 0) 
                return num;
            if(num.charAt(1) == '0'){
                int firstNotZero = 1;
                while(firstNotZero < num.length()  && num.charAt(firstNotZero) == '0') 
                    firstNotZero ++;
                num = num.substring(firstNotZero);
            } else{
                int startIndex = 0;
                while(startIndex < num.length() - 1  && num.charAt(startIndex) <= num.charAt(startIndex + 1)) 
                    startIndex ++;
                num = num.substring(0, startIndex) + num.substring(startIndex + 1);
            }
        }
    }
}