题目：

题解：

class DualHeap {
private:// 大根堆，维护较小的一半元素priority_queue<int> small;// 小根堆，维护较大的一半元素priority_queue<int, vector<int>, greater<int>> large;// 哈希表，记录「延迟删除」的元素，key 为元素，value 为需要删除的次数unordered_map<int, int> delayed;int k;// small 和 large 当前包含的元素个数，需要扣除被「延迟删除」的元素int smallSize, largeSize;public:DualHeap(int _k): k(_k), smallSize(0), largeSize(0) {}private:// 不断地弹出 heap 的堆顶元素，并且更新哈希表template<typename T>void prune(T& heap) {while (!heap.empty()) {int num = heap.top();if (delayed.count(num)) {--delayed[num];if (!delayed[num]) {delayed.erase(num);}heap.pop();}else {break;}}}// 调整 small 和 large 中的元素个数，使得二者的元素个数满足要求void makeBalance() {if (smallSize > largeSize + 1) {// small 比 large 元素多 2 个large.push(small.top());small.pop();--smallSize;++largeSize;// small 堆顶元素被移除，需要进行 pruneprune(small);}else if (smallSize < largeSize) {// large 比 small 元素多 1 个small.push(large.top());large.pop();++smallSize;--largeSize;// large 堆顶元素被移除，需要进行 pruneprune(large);}}public:void insert(int num) {if (small.empty() || num <= small.top()) {small.push(num);++smallSize;}else {large.push(num);++largeSize;}makeBalance();}void erase(int num) {++delayed[num];if (num <= small.top()) {--smallSize;if (num == small.top()) {prune(small);}}else {--largeSize;if (num == large.top()) {prune(large);}}makeBalance();}double getMedian() {return k & 1 ? small.top() : ((double)small.top() + large.top()) / 2;}
};class Solution {
public:vector<double> medianSlidingWindow(vector<int>& nums, int k) {DualHeap dh(k);for (int i = 0; i < k; ++i) {dh.insert(nums[i]);}vector<double> ans = {dh.getMedian()};for (int i = k; i < nums.size(); ++i) {dh.insert(nums[i]);dh.erase(nums[i - k]);ans.push_back(dh.getMedian());}return ans;}
};