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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