Leetcode刷题笔记1 栈与队列part02

leetcode 1047 删除字符串中的所有相邻重复项

class Solution:
    def removeDuplicates(self, s: str) -> str:
        res = list()
        for item in s:
            if res and res[-1] == item:
                res.pop()
            else:
                res.append(item)
        return ''.join(res)

leetcode 150 逆波兰表达式求值

from operator import mul, add, sub
def div(x,y):
    return int(x/y) if x * y > 0 else -(abs(x) // abs(y))
class Solution:
    op_map = {'+': add, '-': sub, '*': mul, '/': div}
    def evalRPN(self, tokens: List[str]) -> int:
        stack = []
        for token in tokens:
            if token not in {'+', '-', '*', '/'}:
                stack.append(int(token))
            else:
                op2 = stack.pop()
                op1 = stack.pop()
                stack.append(self.op_map[token](op1, op2))
        return stack.pop()

 leetcode 239 滑动窗口最大值

滑动窗口这部分比较难理解，需要反复复习

from collections import deque
class Myqueue:
    def __init__(self):
        self.queue = deque()
    def pop(self, value):
        if self.queue and value == self.queue[0]:
            self.queue.popleft()
    def push(self, value):
        while self.queue and value > self.queue[-1]:
            self.queue.pop()
        self.queue.append(value)
    def front(self):
        return self.queue[0]

class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        queue = Myqueue()
        result = []
        for i in range(k):
            queue.push(nums[i])
        result.append(queue.front())
        for i in range(k, len(nums)):
            queue.pop(nums[i - k])
            queue.push(nums[i])
            result.append(queue.front())
        return result

leetcode 347 前k个高频元素

优先级队列（堆排序）

import heapq
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        count_map = {}
        for i in range(len(nums)):
            count_map[nums[i]] = count_map.get(nums[i],0) + 1
        # 对频率进行排序
        pri_que = []
        for key, freq in count_map.items():
            heapq.heappush(pri_que, (freq, key))
            if len(pri_que) > k:
                heapq.heappop(pri_que)
        result = [0] * k 
        for i in range(k-1, -1, -1):
            result[i] = heapq.heappop(pri_que)[1]
        return result

字典解法，有点难

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        time_map = {}
        for i in range(len(nums)):
            time_map[nums[i]] = time_map.get(nums[i], 0) + 1
        index_dict = defaultdict(list)
        for key in time_map:
            index_dict[time_map[key]].append(key)
        key = list(index_dict.keys())
        key.sort()
        result = []
        cnt = 0
        while key and cnt != k:
            result += index_dict[key[-1]]
            cnt += len(index_dict[key[-1]])
            key.pop()
        return result[0:k]