Codility 7.1 Brackets

Determine whether a given string of parentheses (multiple types) is properly nested.

Task description
A string S consisting of N characters is considered to be properly nested if any of the following conditions is true:

S is empty;
S has the form "(U)" or "[U]" or "{U}" where U is a properly nested string;
S has the form "VW" where V and W are properly nested strings.
For example, the string "{[()()]}" is properly nested but "([)()]" is not.

Write a function:

def solution(S)

that, given a string S consisting of N characters, returns 1 if 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.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..200,000];
string S consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".

题目意思：有一个字符串，看是否配对。
思路：用stack 的pop来做。当碰到上括弧时，保存到list中，当碰到下括弧时，就去看与离它最近的那个上括弧是否匹配，这一操作可以由list.pop完成。
注意：1.如果字符串的长度是奇数，返回0
2.注意字符串为空的情况。

def solution(S):
    if len(S) % 2 == 1:   return 0
    matched = {"]":"[", "}":"{", ")": "("}
    to_push = ["[", "{", "("]
    stack = []
    for element in S:
        if element in to_push:
            stack.append(element)
        else:
            if len(stack) == 0:
                return 0
            elif matched[element] != stack.pop():
                return 0
    if len(stack) == 0:
        return 1
    else:
        return 0