一、题目说明
题目239. Sliding Window Maximum,给一个数组,和窗口长度,窗口每次向右滑动1位,返回滑动窗口的最大值。时间复杂度要求是线性。
二、我的解答
最直接的办法就是brute force
,但是性能不足,复杂度是O(kN):
class Solution{
public:
//brute force
vector maxSlidingWindow(vector& nums, int k){
int start = 0,curMax=0;
vector result;
while(start+k <= nums.size()){
curMax = nums[start];
for(int i=start;icurMax) curMax = nums[i];
}
result.push_back(curMax);
start++;
}
return result;
}
};
性能如下:
Runtime: 100 ms, faster than 16.44% of C++ online submissions for Sliding Window Maximum.
Memory Usage: 13 MB, less than 86.89% of C++ online submissions for Sliding Window Maximum.
三、优化措施
用双端队列实现:
class Solution{
public:
//use deque to store index,window.front() to store the max value of current window
//nums[i],将双端队列deque从尾部开始把每个小于nums[i]的数去除。
//nums[i-k]等于deque头的元素,要把deque的头去除掉,相当于一个最大值移出了窗口
vector maxSlidingWindow(vector& nums, int k){
vector result;
if(k==0 || nums.size()<1) return result;
deque window;
//init the first k nums
for(int i=0;inums[window.back()]){
window.pop_back();
}
window.push_back(i);
}
result.push_back(nums[window.front()]);
//other data
for(int i=k;inums[window.back()]){
window.pop_back();
}
window.push_back(i);
result.push_back(nums[window.front()]);
}
return result;
}
};
性能如下:
Runtime: 60 ms, faster than 63.43% of C++ online submissions for Sliding Window Maximum.
Memory Usage: 13.3 MB, less than 59.02% of C++ online submissions for Sliding Window Maximum.