核心思路使用栈和map抓住右括号匹配 时间复杂度o(n) 需注意点出栈时需要判空 def search(nums): if not nums: return True stack [] m {]: [, }: {, ): (} for i in nums: if i not in m: stack.append(i) else: k stack.pop() if stack else if k ! m[i]: return False return not stack if __name__ __main__: s {[()]} result search(s) print(result)