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# 402    Remove K Digits

## 402 [Remove K Digits](https://leetcode.com/problems/remove-k-digits/description/)

## 1. Question

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.
```

## 2. Implementation

思路：这题思路和Remove Duplicate Letters一样，利用单调栈保持输出结果的单调性，只是要注意一些Corner cases

```java
class Solution {
    public String removeKdigits(String num, int k) {
        int len = num.length();

        if (k == len) {
            return "0";
        }

        Stack<Character> stack = new Stack<>();

        for (char digit : num.toCharArray()) {
            while (k > 0 && !stack.isEmpty() && stack.peek() > digit) {
                stack.pop();
                --k;
            }

            stack.push(digit);
        }

        while (k > 0) {
            stack.pop();
            --k;
        }

        StringBuilder res = new StringBuilder();
        while (!stack.isEmpty()) {
            res.append(stack.pop());
        }

        while (res.length() > 1 && res.charAt(res.length() - 1) == '0') {
            res.deleteCharAt(res.length() - 1);
        }
        return res.reverse().toString();
    }
}
```

## 3. Time & Space Complexity

时间和空间复杂度都是O(n)