Leetcode brushing record-402. Remove K digits

I have implemented two methods:
1. The brute force method.
First, consider one input and only delete 1 number to minimize the result.
Then, for an input num and k, we perform the above operations on num k times, then

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        length = len(num)
        num_val = int(num)

        if num_val == 0 or length <= k:
            return "0"
        num_list = []
        for letter in num:
            num_list.append(int(letter))
        print(num_list)

        #num_list
        start_index = 0
        for i in range(k):#执行k次
            length = len(num_list)
            for j in range(length-1):
                if num_list[j] > num_list[j+1]:
                    num_list.pop(j)
                    break
                if j == length - 2 and num_list[j] <= num_list[j+1]:
                    num_list.pop(-1)
                    start_index
        newlist = num_list
        finallength = len(newlist)
        if list(set(newlist)) == [0]:
            return "0"
        index = 0
        while newlist[index] == 0:
            index += 1
        res = ''
        for value in newlist[index:]:
            res += str(value)
        return res

2. Stack operation based on greedy algorithm

We first build a stack to store the reserved numbers.
Calculate index = len (array)-k

Iterate over each number in the array:
For the first number, put it on the stack first
For each subsequent:

If the number to be deleted k == 0:
append the following numbers in the array to the stack

If the number to be deleted k> 0 and the number being traversed by the array <stack top:
while k> 0 and stack non-empty and stack top> this number:
stack top pop
k-= 1 After
jumping out of the loop, the
stack .append (this number)

If the number to be deleted k> 0 and the number that the array is traversing> = stack top:
stack.append (this number)

After jumping out of the traversal,
if k> 0:
only take the front of the stack [: index]

If there is only 0 in the stack:
return str (0)

If the stack [0] == 0
gets the first number not equal to 0, from it to the stack [-1], returns a str

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        length = len(num)
        num_val = int(num)
        #oldk = k
        if num_val == 0 or length <= k:
            return "0"
        num_list = []
        for letter in num:
            num_list.append(int(letter))
        print(num_list)

        should = length - k
        #num_list
        zhan = []
        #zhan#.append(num_list[0])
        tempindex = 0
        shanwan = False
        for i in range(length):
            if k == 0:
                for value in num_list[i:]:  
                    zhan.append(value)#shanwan = True
                break
            elif zhan == []:
                zhan.append(num_list[i])
            else:
                if zhan[-1] <= num_list[i]:
                    zhan.append(num_list[i])

                elif zhan[-1] > num_list[i]:
                    while k > 0 and zhan != [] and zhan[-1] > num_list[i]:
                        zhan.pop(-1)
                        k -= 1
                    zhan.append(num_list[i])
        if k > 0:
            zhan = zhan[:should ]
        print(zhan)
        #while len(zhan) < should:
        #    zhan.append(num_list[tempindex + 1])
        #    tempindex += 1
        if list(set(zhan)) == [0]:#0
            return '0'
        while zhan[0] == 0:
            zhan.pop(0)
        res = ''
        for value in zhan:
            res += str(value)
        return res