有效的括号

def solve():
    s = '()[]{}'
    stack = []
    mapping = {')': '(', ']': '[', '}': '{'}
    valid = True
    for char in s:
        if char in mapping:
            if not stack or stack.pop() != mapping[char]:
                valid = False
                break
        else:
            stack.append(char)
    print(valid and not stack)