402. Remove K Digits

Medium (Trung bình) C++ Python 🔗 Xem trên LeetCode

📋 Đề Bài

Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num.

 

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.

 

Constraints:

  • 1 <= k <= num.length <= 105
  • num consists of only digits.
  • num does not have any leading zeros except for the zero itself.

🧠 Thuật Toán & Kỹ Thuật

Stack (Ngăn xếp)String (Chuỗi)
⏱️ Thời gian O(n²)
💾 Không gian O(1)

💻 Lời Giải

C++ 0402-remove-k-digits.cpp
class Solution {
public:
    string removeKdigits(string num, int k) {
        int n = num.size();
        if (n <= k) {
            return "0";
        }
        if (k == 0) {
            return num;
        }
        stack<char> st;
        st.push(num[0]);
        for (int i = 1; i < n; ++i) {
            while (k > 0 and !st.empty() and st.top() > num[i]) {
                st.pop();
                --k;
            }
            st.push(num[i]);
            if (st.size() == 1 and num[i] == '0') {
                st.pop();
            }
        }
        while (k > 0 and !st.empty()) {
            st.pop();
            --k;
        }
        string res;
        while (!st.empty()) {
            res.push_back(st.top());
            st.pop();
        }
        if (res.size() == 0) {
            return "0";
        }
        reverse(res.begin(), res.end());
        return res;
    }
};
Python 0402-remove-k-digits.py
class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        stack = []
        
        for d in num:
            while stack and stack[-1] > d and k > 0:
                stack.pop()
                k -= 1
            stack.append(d)
            
        while k > 0:
            stack.pop()
            k -= 1
            
        i = 0
        
        while i < len(stack) and stack[i] == '0':
            i += 1
            
        res = "".join(stack[i:])
        
        return res or "0"