239.Sliding Window Maximum¶
Tags: Hard Heap Sliding Window
Links: https://leetcode.com/problems/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.
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
Note: You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.
Follow up: Could you solve it in linear time?
解法一：ST表的RMQ¶
class Solution {
int d[100005][21];
int n;
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
if (nums.size() == 0) return {};
n = nums.size();
vector<int> res(n - k + 1);
init(nums);
for (int i = 0; i + k - 1 < n; ++i) {
res[i] = RMQ(i + 1, i + k);
}
return res;
}
void init(const vector<int> & nums)
{
for (int i = 1; i <= n; ++i) d[i][0] = nums[i - 1];
for (int j = 1; (1 << j) <= n; ++j) {
for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
d[i][j] = max(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
}
}
int RMQ(int L, int R)
{
int k = log2(R - L + 1);
return max(d[L][k], d[R - (1 << k) + 1][k]);
}
};
很典型的RMQ问题，采用ST表，已经在《数据结构——ST表》总结过了，建表的时间复杂度为O(n \log n)，查询操作为O(1)。
如果要求线性时间复杂度，其实就是利用线段树来进行建表操作，建表时间复杂度为O(n)，查询为\log n，时间复杂度仍然是O(n \log n)。
解法二：单调队列¶
其实很明显这个就是**单调队列**的模板题目。在《挑战程序设计竞赛》的4.4.2节
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
vector<int> res;
deque<int> q;
for (int i = 0; i < nums.size(); ++i) {
if (!q.empty() && q.front() == i - k) q.pop_front();
while (!q.empty() && nums[q.back()] < nums[i]) q.pop_back();
q.push_back(i);
if (i >= k - 1) res.push_back(nums[q.front()]);
}
return res;
}
};
Runtime: 44 ms, faster than 99.94% of C++ online submissions for Sliding Window Maximum.
Memory Usage: 12.3 MB, less than 100.00% of C++ online submissions for Sliding Window Maximum.
解法三：区间树¶
《算法问题实战策略》17.4其他学习内容，区间树可以快速求解区间乘积、区间最小/最大值、最大出现频率等（莫队算法也可以解决出现频率问题）