原题

Given an array 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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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

Therefore, return the max sliding window as [3,3,5,5,6,7].

Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Follow up:
Could you solve it in linear time?

地址

https://leetcode.com/problems/sliding-window-maximum/

程序范例

[code lang=”cpp”]
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int rLength = nums.size() – k + 1;
if (nums.size() == 0) {
vector<int> r(0,0);
return r;
}
vector<int> r(rLength,-999999999);
vector<int>::iterator item = nums.begin();
int index = 0;
while(item != nums.end()) {
for (int j = index; (j >= 0 && j >= index + 1 – k); j–) {
if (j >= rLength) {
continue;
}
r[j] = max(r[j], *item);
}
++index;
++item;
}
return r;
}
};
[/code]

程序运行分布图

看来，大部分的运行时间在80ms左右，而上述代码的运行时间却在500ms，必须优化。看原题提示：

 How about using a data structure such as deque (double-ended queue)?

未完待续。。。

文章来源：胡旭博客

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