题目
You are given a string s and an integer k, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them, causing the left and the right side of the deleted substring to concatenate together.
We repeatedly make k duplicate removals on s until we no longer can.
Return the final string after all such duplicate removals have been made. It is guaranteed that the answer is unique.
解题思路
使用栈，栈里元素为当前字母连续出现了几次，到k次就pop
代码
class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = [['#', 0]]
for c in s:
if c == stack[-1][0]:
stack[-1][1] += 1
if stack[-1][1] == k:
stack.pop()
else:
stack.append([c, 1])
return ''.join([c[0] * c[1] for c in stack[1:]])
