[leetcode] 402. Remove K Digits @ python

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原题

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.

解法

堆栈+贪心算法. 构造一个stack, 储存一个递增的数列, 遍历num, 如果当前数字大于堆栈中最后一个数字, 那么一直出栈, 直到当前数字放入之后stack是递增的数列. 如果遍历完成之后k个数字还没用完, 那么从stack最右边的数字开始删除, 直到K为0.

Time: O(n)
Space: O(n)

代码

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        # edge case
        if len(num) <= k: return '0'
        
        stack = []
        for i, digit in enumerate(num):
            while k > 0 and stack and int(stack[-1]) > int(digit):
                # remove the digit
                stack.pop()
                k -= 1
            stack.append(digit)
        
        while k > 0:
            stack.pop()
            k -= 1
            
        ans = ''.join(stack).lstrip('0')
        return ans if ans else '0'