Leetcode: 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.

Greedy + Stack: 用一个栈维护最后要留存下来的digits

需要注意的是：如果遍历到string的某个char, string后面的char数刚刚好能填满这个栈，那么即使这个char比栈顶还要小，也不出栈，直接入栈

最后要删除leading 0

public class Solution {
    public String removeKdigits(String num, int k) {
        if (num.length()==0 || k>=num.length()) return "0";
        char[] arr = num.toCharArray();
        Stack<Character> stack = new Stack<Character>();
        StringBuilder res = new StringBuilder();
        int size = arr.length - k;
        for (int i=0; i<arr.length; i++) {
            while (!stack.isEmpty() && arr[i]<stack.peek() && (size-stack.size()+1 <= arr.length-i)) {
                stack.pop();
            }
            if (size > stack.size())
                stack.push(arr[i]);
        }
        while (!stack.isEmpty()) {
            res.insert(0, stack.pop());
        }
        while (res.length()>1 && res.charAt(0)=='0') res.deleteCharAt(0);
        return res.toString();
    }
}

用char[]实现栈，思路一样，要快很多，beat96%

public class Solution {
    public String removeKdigits(String num, int k) {
        int remain = num.length() - k;
        char[] numArray = num.toCharArray(), res = new char[remain];
        int index = 0;
        for(int i = 0; i < numArray.length; i++) {
            while((numArray.length - i > remain - index) && (index > 0 && numArray[i] < res[index - 1])) index--;
            if(index < remain) res[index++] = numArray[i];
        }
        
        // check leading zeroes
        index = -1;
        while(++index < remain) {
            if(res[index] != '0') break;
        }
        String s = new String(res).substring(index);
        
        return s.length() == 0 ? "0" : s;
    }
}