Sliding Window Maximum
239. Sliding Window Maximum
You are given an array of integers nums
, there is a sliding window of size k
which is moving from the very left of the array to the very right. You can only see the k
numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
--------------- -----
[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
Example 2:
Input: nums = [1], k = 1
Output: [1]
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
1 <= k <= nums.length
Analysis
We can use the monotonic stack to resolve the problem. We use a stack named s
which store the unresolved index,
- we compare the current height
height[i]
with the top of stacks[-1]
. Ifheight[i] > s[-1]
, it means we cound find a target trap, we can pop the index from the stack twice and calculate the trap area; then continue the next round until one of bellow condition meet:- The stack is empty
- The height of stack’s top is bigger than
height[i]
- If it’s less, then we should just push the current index.
Solutions
1. Monotanic-Stack
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
s = []
n = len(nums)
ans = 0
q = deque()
for i in range(n):
while q and q[-1] <= nums[i]:
q.pop()
q.append(i)
if i - q[0] >= k:
q.popleft()
if i > k:
ans.append(nums[q[0]])
return ans
Input: [0,1,0,2,1,0,1,3,2,1,2,1]
output:
6
Complexity
- Time complexity: ( O(n) ), where ( n ) is the length of the array.
- Space complexity: ( O(n) ), where ( n) is the length of the array.
链接:https://ac.nowcoder.com/acm/contest/1006/D 来源:牛客网
题目描述
输入一个长度为n的整数序列,从中找出一段不超过m的连续子序列,使得整个序列的和最大。 例如 1,-3,5,1,-2,3 当m=4时,S=5+1-2+3=7 当m=2或m=3时,S=5+1=6
输入描述:
第一行两个数n,m(n,m≤300000)(n,m \leq 300000)(n,m≤300000)
第二行有n个数,要求在n个数找到最大子序和
输出描述:
一个数,数出他们的最大子序和
示例1
输入
6 4
1 -3 5 1 -2 3
输出
7
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