有效的括号
给定一个只包括'(',')','{','}','[',']'的字符串s,判断字符串是否有效。
有效字符串需满足:
- 左括号必须用相同类型的右括号闭合。
- 左括号必须以正确的顺序闭合。
- 每个右括号都有一个对应的相同类型的左括号。
示例 1:
输入:s = "()"
输出:true
示例 2:
输入:s = "()[]{}"
输出:true
示例 3:
输入:s = "(]"
输出:false
示例 4:
输入:s = "([])"
输出:true
示例 5:
输入:s = "([)]"
输出:false
提示:
1 <= s.length <= 104s仅由括号'()[]{}'组成
class Solution: def isValid(self, s: str) -> bool: if len(s) % 2 == 1: return False pairs = { ")": "(", "]": "[", "}": "{", } stack = list() for ch in s: if ch in pairs: if not stack or stack[-1] != pairs[ch]: return False stack.pop() else: stack.append(ch) return not stack最小栈
设计一个支持push,pop,top操作,并能在常数时间内检索到最小元素的栈。
实现MinStack类:
MinStack()初始化堆栈对象。void push(int val)将元素val推入堆栈。void pop()删除堆栈顶部的元素。int top()获取堆栈顶部的元素。int getMin()获取堆栈中的最小元素。
示例 1:
输入:["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]]输出:[null,null,null,null,-3,null,0,-2]解释:MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.getMin(); --> 返回 -2.
提示:
-231 <= val <= 231 - 1pop、top和getMin操作总是在非空栈上调用push,pop,top, andgetMin最多被调用3 * 104次
借用了一个辅助栈min_stack[ ]
class MinStack: def __init__(self): self.stack = [] self.min_stack = [math.inf] def push(self, val: int) -> None: self.stack.append(val) self.min_stack.append(min(val, self.min_stack[-1])) def pop(self) -> None: self.stack.pop() self.min_stack.pop() def top(self) -> int: return self.stack[-1] def getMin(self) -> int: return self.min_stack[-1] # Your MinStack object will be instantiated and called as such: # obj = MinStack() # obj.push(val) # obj.pop() # param_3 = obj.top() # param_4 = obj.getMin()