【Python】Codility in Python : Lesson 7 - Stacks and Queues【Nesting】

Stacks and Queues 第四題：【Nesting】
Determine whether given string of parentheses is properly nested.

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A string S consisting of N characters is called properly nested if:

S is empty;
S has the form "(U)" where U is a properly nested string;
S has the form "VW" where V and W are properly nested strings.
For example, string "(()(())())" is properly nested but string "())" isn't.

Write a function:

　　def solution(S)

that, given a string S consisting of N characters,
returns 1 if string S is properly nested and 0 otherwise.

For example, given S = "(()(())())", the function should return 1 and given S = "())",
the function should return 0, as explained above.

Assume that:

　　N is an integer within the range [0..1,000,000];
　　string S consists only of the characters "(" and/or ")".

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方法一：用 append 和 pop 實作 stack 的概念
Correctness：100%、Performance：100%

def solution(S):
  stack = []
  for sign in S:
    if sign == '(':
      stack.append(sign)
    else:
      if len(stack) == 0:
        return 0
      stackLast = stack[-1]
      if sign == ')' and stackLast == '(':
        stack.pop()
      else:
        return 0

  return 1 if len(stack) == 0 else 0

完整練習題 source code 請參閱：github