6. 스택

컨트롤 제트 - https://school.programmers.co.kr/tryouts/85926/challenges

리스트에서 pop과 append로 스택 구현 가능

def solution(s):
    answer = 0
    stack = []
    s_list = s.split(" ")
    for item in s_list:
        if item == "Z":
            stack.pop()
        else:
            stack.append(int(item))
    return sum(stack)

 

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같은 숫자는 싫어 - https://school.programmers.co.kr/tryouts/85927/challenges

직전 숫자만 기억

def solution(arr):
    ans = []
    before = -1
    for num in arr:
        if num != before:
            ans.append(num)
            before = num
    return ans

 

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햄버거 만들기 - https://school.programmers.co.kr/tryouts/85928/challenges

def solution(ingredient):
    cnt = 0
    stack = []
    for num in ingredient:
        if num == 1 and stack[-3:] == [1,2,3]: # 마지막 빵 들어올 때
            for _ in range(3):
                stack.pop()
            cnt += 1
        else:
            stack.append(num)
    return cnt

 

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올바른 괄호 - https://school.programmers.co.kr/tryouts/85929/challenges

def solution(s):
    answer = True
    stack = []
    for c in s:
        if c == "(":
            stack.append(1)
        else:
            if len(stack) == 0:
                return False
            stack.pop()
    if len(stack) != 0:
        return False
    return True