leetcode402. 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.

10200 k=1
第一步： 0和1比较，此时0比1小，且还有一次可用的删除，因此删除1，保留0
第二步：因为无可用的删除次数，因此剩下的值全部保留

123450123 k=5
第一步：2>1 保留
第二步：3>2 保留
第三步: 4>3 保留
第四步: 5>4 保留
第五步：0<5 删除5 保留0 k=4
第六步: 0<4 删除4 保留0 k=3
第七步：0<3 删除3 保留0 k=2
第八步：0<2 删除2 保留0 k=1
第九步：0<1 删除1 保留0 k=0
第十步: 此时所有的删除次数用完，因此剩余的值全部保留

    public String removeKdigits(String num, int k) {
        if(num == null || num.length() == 0 || k==num.length()) return "0";
        Stack<Character> stack = new Stack<>();
        for(int i = 0 ; i < num.length() ; i++) {
            char c = num.charAt(i);
            while(k> 0 && !stack.isEmpty() && stack.peek() > c) {
                stack.pop();
                k--;
            }
            stack.push(c);
        }
        
        while(k > 0) {
            stack.pop();
            k--;
        }
        
        StringBuilder result = new StringBuilder();
        while(!stack.isEmpty()) {
            result.append(stack.pop());
        }
        while(result.length() > 1 && result.charAt(result.length()-1) == '0') {
            result.deleteCharAt(result.length()-1);
        }
        return result.reverse().toString();
    }