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.

找出去掉k个数后的所组成的最小数值。

思路一：通过回溯算法，找出len-k个数组成的最小数，最后一个用例会TLE，程序如下：

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class Solution {
    int min = Integer.MAX_VALUE, cur = 0;
    public String removeKdigits(String num, int k) {
        backTracing(num, num.length(), num.length() - k, 0);
        return String.valueOf(min);
    }
    
    public void backTracing(String num, int len, int k, int begin){
        if (begin > len){
            return;
        }
        if (k == 0){
            min = Math.min(min, cur);
            return;
        }
        for (int i = begin; i < len; ++ i){
            cur = cur*10 + num.charAt(i) - '0';
            backTracing(num, len, k - 1, i + 1);
            cur = (cur - num.charAt(i) + '0')/10;
        }
    }
}

思路2：通过DP求解，利用栈来实现，程序如下所示：

class Solution {
    public String removeKdigits(String num, int k) {
        ArrayDeque<Character> stack = new ArrayDeque<>();
        int len = num.length();
        for (int i = 0; i < len; ++ i){
            while (k > 0 && !stack.isEmpty() && stack.peek() > num.charAt(i)){
                stack.pop();
                k --;
            }
            stack.push(num.charAt(i));
        }
        while (k > 0){
            stack.pop();
            k --;
        }
        StringBuilder sb = new StringBuilder();
        while (!stack.isEmpty()){
            sb.append(stack.pop());
        }
        String s = sb.reverse().toString();
        len = s.length();
        for (int i = 0; i < len; ++ i){
            if (s.charAt(i) != '0'){
                return s.substring(i, len);
            }
        }
        return "0";
    }
}

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