動機

讓前面數字越小越好!!

Problem

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 = 3Output: "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 = 1Output: "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 = 2Output: "0"Explanation: Remove all the digits from the number and it is left with nothing which is 0.

 

Constraints:

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

Sol

在保持順序的狀態下，維持某一條件，Monotonic Stack

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        stk = []
        for c in num:
            while stk and k and stk[-1] > c:
                stk.pop()
                k -= 1
            stk.append(c)
        for _ in range(k):
            if stk:
                stk.pop()
        i = 0
        while i < len(stk) and stk[i] == '0':
            i += 1
        stk= stk[i:]
        
        return "0" if not stk else ''.join(stk)