1. 滑动窗口最大值
给你一个整数数组 nums
,有一个大小为 k
的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k
个数字。滑动窗口每次只向右移动一位。
返回滑动窗口中的最大值。
示例 1:
输入:nums = [1,3,-1,-3,5,3,6,7], k = 3
输出:[3,3,5,5,6,7]
解释:
滑动窗口的位置 最大值
--------------- -----
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
示例 2:
输入:nums = [1], k = 1
输出:[1]
示例 3:
输入:nums = [1,-1], k = 1
输出:[1,-1]
示例 4:
输入:nums = [9,11], k = 2
输出:[11]
示例 5:
输入:nums = [4,-2], k = 2
输出:[4]
提示:
1 <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4
1 <= k <= nums.length
出处:
https://edu.csdn.net/practice/26377425
代码:
class Solution(object): def maxSlidingWindow(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ if len(nums) == 0: return [] if k == 1: return nums res = [] res.append(max(nums[:k])) for i in range(1, len(nums) - k + 1): m = max(nums[i : i + k]) res.append(m) return res if __name__ == '__main__': s = Solution() nums = [1,3,-1,-3,5,3,6,7]; k = 3 print(s.maxSlidingWindow(nums, k))
输出:
[3, 3, 5, 5, 6, 7]
2. 用栈实现队列
请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作(push
、pop
、peek
、empty
):
实现 MyQueue
类:
void push(int x)
将元素 x 推到队列的末尾int pop()
从队列的开头移除并返回元素int peek()
返回队列开头的元素boolean empty()
如果队列为空,返回true
;否则,返回false
说明:
- 你只能使用标准的栈操作 —— 也就是只有
push to top
,peek/pop from top
,size
, 和is empty
操作是合法的。 - 你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。
进阶:
- 你能否实现每个操作均摊时间复杂度为
O(1)
的队列?换句话说,执行n
个操作的总时间复杂度为O(n)
,即使其中一个操作可能花费较长时间。
示例:
输入:
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
输出:
[null, null, null, 1, 1, false]
解释:
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false
提示:
1 <= x <= 9
- 最多调用
100
次push
、pop
、peek
和empty
- 假设所有操作都是有效的 (例如,一个空的队列不会调用
pop
或者peek
操作)
出处:
https://edu.csdn.net/practice/26377426
代码:
class MyQueue: def __init__(self): """ Initialize your data structure here. """ self.stack_push = [] self.stack_pop = [] def push(self, x: int) -> None: """ Push element x to the back of queue. """ while self.stack_pop: self.stack_push.append(self.stack_pop.pop()) self.stack_push.append(x) def pop(self) -> int: """ Removes the element from in front of queue and returns that element. """ if self.empty(): return None else: while self.stack_push: self.stack_pop.append(self.stack_push.pop()) return self.stack_pop.pop() def peek(self) -> int: """ Get the front element. """ if self.empty(): return None elif not self.stack_pop: return self.stack_push[0] else: return self.stack_pop[-1] def empty(self) -> bool: """ Returns whether the queue is empty. """ return not (self.stack_push or self.stack_pop) # Your MyQueue object will be instantiated and called as such: # obj = MyQueue() # obj.push(x) # param_2 = obj.pop() # param_3 = obj.peek() # param_4 = obj.empty() myQueue = MyQueue() myQueue.push(1) # queue is: [1] myQueue.push(2) # queue is: [1, 2] (leftmost is front of the queue) print(myQueue.peek()) # return 1 print(myQueue.pop()) # return 1, queue is [2] print(myQueue.empty()) # return false
输出:
1
1
False
3. 直线上最多的点数
给你一个数组 points
,其中 points[i] = [xi, yi]
表示 X-Y 平面上的一个点。求最多有多少个点在同一条直线上。
示例 1:
输入:points = [[1,1],[2,2],[3,3]]
输出:3
示例 2:
输入:points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
输出:4
提示:
1 <= points.length <= 300
points[i].length == 2
-104 <= xi, yi <= 104
points
中的所有点 互不相同
出处:
https://edu.csdn.net/practice/26377427
代码:
from typing import List class Solution: def maxPoints(self, points: List[List[int]]) -> int: if len(points) <= 2: return len(points) maxNum = 0 for i in range(len(points)): maxNum = max(maxNum, self.find(points[i], points[:i] + points[i + 1 :])) return maxNum + 1 def find(self, point, pointList): memories = {} count = 0 inf = float("inf") for curPoint in pointList: if curPoint[1] == point[1] and curPoint[0] != point[0]: memories[inf] = memories.get(inf, 0) + 1 elif curPoint[1] == point[1] and curPoint[0] == point[0]: count += 1 else: slope = (curPoint[0] - point[0]) / (curPoint[1] - point[1]) memories[slope] = memories.get(slope, 0) + 1 if memories: return count + max(memories.values()) else: return count if __name__ == '__main__': s = Solution() null = None points = [[1,1],[2,2],[3,3]] print(s.maxPoints(points)) points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] print(s.maxPoints(points))
输出:
3
4
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