- Sliding Window Maximum
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. Return the max sliding window.
Follow up:
Could you solve it in linear time?
Example:
Input: nums = [1,3,-1,-3,5,3,6,7], and 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
解 很奇怪很多答案为什么又是堆又是哈希表的。。。
用一个tmp_max变量记录窗口里的最大值，向右移动时，如果最左边出去的是最大值tmp_max，就在窗口里重新线性搜索或者调用max_element()或者其他什么方法找到最大值，右端点进来的新的值和tmp_max比较一下取大的作为新的tmp_max
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int l = 0, r = k;
vector<int>ans;
int tmp = INT_MIN;
for(int i = 0; i < r; ++i)tmp = max(nums[i], tmp);
ans.push_back(tmp);
r++;
l++;
while(r <= nums.size()){
if(tmp == nums[l-1]){
tmp = INT_MIN;
for(int i = l; i < r; ++i)tmp = max(tmp, nums[i]);
}else{
tmp = max(tmp, nums[r-1]);
}
ans.push_back(tmp);
l++;
r++;
}
return ans;
}
};