LeetCode-Microsoft-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：其基本思想是利用栈尽量维持一个递增的序列，也就是说将字符串中字符依次入栈，如果当前字符串比栈顶元素小，并且还可以继续删除元素，那么就将栈顶元素删掉，这样可以保证将当前元素加进去一定可以得到一个较小的序列．也可以算是一个贪心思想．最后我们只取前len-k个元素构成一个序列即可，如果这样得到的是一个空串那就手动返回０．还有一个需要注意的是字符串首字符不为０

class Solution {
    public String removeKdigits(String num, int k) {
        if(num == null || num.length() == 0){
            return null;
        }
        Stack<Integer> stack = new Stack<>();

        for(int i = 0; i<num.length(); i++){
            int cur = num.charAt(i) - ‘0‘;
            while(!stack.isEmpty() && cur < stack.peek() && num.length() - i - 1 >= num.length()-k-stack.size()){
                stack.pop();
            }
            if(stack.size() < num.length()-k){
                stack.push(cur);
            }
        }
        StringBuilder res = new StringBuilder();
        int count = 0;
        while (!stack.isEmpty()){
            res.insert(0, stack.pop());
        }

        while (res.length() > 0 && res.charAt(0) == ‘0‘){
            res.deleteCharAt(0);
        }

        if(res.length() == 0){
            return "0";
        }
        return res.toString();
    }
}

原文地址：https://www.cnblogs.com/incrediblechangshuo/p/8970525.html