题目描述:

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:

1
2
3
4
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:

1
2
3
4
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:

1
2
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.

这道题的主要思想就是"如果遇到了一个较小的值, 那么就把已得到的字符串中尾部所有比它大的值删除, 再把它放入末尾; 如果比末尾元素大, 那么就把它直接添加到末尾". 这是一个栈的问题, 但是除了主要的思路以外还有另一个问题: 最终栈的大小必须是num.size()-k, 所以在添加和删除时必须要考虑栈的大小.

在元素比栈顶元素大的情况下还要增加一个条件: 栈的大小还没有达到num.size()-k.

而元素比栈顶元素小的情况下, 在弹出栈顶元素时要求num中剩下的元素数量>=栈的目标大小-栈目前的大小, 这样才能保证栈能被填满.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution {
public:
    string removeKdigits(string num, int k) {
        string ans;
        int target = num.size() - k, size = num.size(), i;
        if(target == 0) return "0";
        ans.push_back(num[0]);
        for(i = 1; i < k + ans.size(); i++){
            if(num[i] > ans.back() && ans.size() < target) ans.push_back(num[i]);
            else{
                while(!ans.empty() && ans.back() > num[i] && size - i - 1 >= target - ans.size()) ans.pop_back();
                if(ans.size() < target) ans.push_back(num[i]);
            }
        }
        for(; ans.size() < target; i++){
            ans.push_back(num[i]);
        }
        for(i = 0; ans[i] == '0'; i++);
        if(ans.empty() || i == ans.size()) return "0";
        return ans.substr(i);
    }
};