402. Remove K Digits.md

402. Remove K Digits

https://leetcode.com/problems/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

• 用单调栈维护一个递增的序列，发现当前数字比栈顶元素要小的数字就 pop 栈顶
• 如果遍历完整个 string 还有多余的 k(k>0)，说明还要再继续 pop，则只需要从栈顶 pop 掉 k 个元素
• 还要处理 leading zero 的情况，只需要从 0 开始遍历栈，若 stack[i] = "0"，i++
• 返回处理完 leading zero 之后的栈，如果栈空，那么返回0，表示我们已经 remove 完所有的元素。

代码

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        
        stack = []
        for x in num: 
            while stack and int(stack[-1]) > int(x) and k:
                stack.pop()
                k -= 1
            stack.append(str(x))
        
        while k:
            # all is increasing
            stack.pop()
            k -= 1
        
        i = 0
        while i < len(stack) and stack[i] == "0":
            # skip leading zero
            i += 1
        
        return "".join(stack[i:]) if len(stack[i:]) > 0 else "0"