2017年2月22日 星期三

【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.
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 ")".


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

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