leetcode(2)栈 - 技术栈

leetcode 155 最小栈

stack相当于栈，先进后出存储全部栈元素 [-3,2,-1]

min_stack,存储栈当前位置最小的元素 [-3,-3,-3]

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class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = [math.inf]

    def push(self, x: int) :
        self.stack.append(x)
        self.min_stack.append(min(x, self.min_stack[-1]))

    def pop(self) -> None:
        self.stack.pop()
        self.min_stack.pop() #  也需要删除当前位置的最小元素

    def top(self) -> int:
        return self.stack[-1]

    def getMin(self) -> int:
        return self.min_stack[-1]

leetcode 20 有效括号

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class Solution:
    def isValid(self, s: str) -> bool:
        dic = {'{': '}',  '[': ']', '(': ')','?':'?'}
        stack = ["?"]
        for c in s:
            if c in dic: stack.append(c)
            elif dic[stack.pop()] != c: return False 
        return len(stack) == 1

leetcode 227 基本计算器2

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class Solution:
    def calculate(self, s: str) -> int:
        n = len(s)
        stack = []
        preSign = '+'
        num = 0
        for i in range(n):
            if s[i] != ' ' and s[i].isdigit():
                num = num * 10 + ord(s[i]) - ord('0')
            if i == n - 1 or s[i] in '+-*/':
                if preSign == '+':
                    stack.append(num)
                elif preSign == '-':
                    stack.append(-num)
                elif preSign == '*':
                    stack.append(stack.pop() * num)
                else:
                    stack.append(int(stack.pop() / num))
                preSign = s[i]
                num = 0
        return sum(stack)

leetcode 150 逆波兰表达式

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class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        op_to_binary_fn = {
            "+": add,
            "-": sub,
            "*": mul,
            "/": lambda x, y: int(x / y),   # 需要注意 python 中负数除法的表现与题目不一致
        }

        stack = list()
        for token in tokens:
            try:
                num = int(token)
            except ValueError:
                num2 = stack.pop()
                num1 = stack.pop()
                num = op_to_binary_fn[token](num1, num2)
            finally:
                stack.append(num)
            
        return stack[0]

leetcode 验证栈序列

输入两个栈 A和B判断B是否为A的出栈序列

输入：pushed = [1,2,3,4,5], popped = [4,5,3,2,1]

输出：true

解释：我们可以按以下顺序执行：

push(1), push(2), push(3), push(4), pop() -> 4,

push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1

复制代码

class Solution:
    def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
        stack, i = [], 0
        for num in pushed:
            stack.append(num) # num 入栈
            while stack and stack[-1] == popped[i]: # 循环判断与出栈
                stack.pop()
                i += 1
        return not stack

leetcode 字符串编解码

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class Solution:
    def decodeString(self, s: str) -> str:
        stack, res, multi = [], "", 0
        for c in s:
            if c == '[':
                stack.append([multi, res])
                res, multi = "", 0
            elif c == ']':
                cur_multi, last_res = stack.pop()
                res = last_res + cur_multi * res
            elif '0' <= c <= '9':
                multi = multi * 10 + int(c)            
            else:
                res += c
        return res

leetcode 下一个更大元素

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class Solution1:
    def nextGreaterElement(self, nums1, nums2):
        dic = {}
        for i in range(len(nums2)):
            j = i + 1
            while j < len(nums2) and nums2[i] >= nums2[j]:
                j += 1
            if j < len(nums2) and nums2[i] < nums2[j]:
                dic[nums2[i]] = nums2[j]
        return [dic.get(x, -1) for x in nums1]

leetcode 去除重复字符

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class Solution(object):
    def removeKdigits(self, num, k):
        stack = []
        remain = len(num) - k
        for digit in num:
            while k and stack and stack[-1] > digit:
                stack.pop()
                k -= 1
            stack.append(digit)
        return ''.join(stack[:remain]).lstrip('0') or '0'

leetcode 每日最高温度

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class Solution:
    def dailyTemperatures(self, temperatures):
        stack, ret = [], [0] * len(temperatures)
        for i, num in enumerate(temperatures):
            while stack and temperatures[stack[-1]] < num:
                index = stack.pop()
                ret[index] = i - index
            stack.append(i)
        return ret