leetcode 480. Sliding Window Median 滑动窗口中位数

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

Examples:
[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

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. Your job is to output the median array for each window in the original array.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position Median
————— —–
[1 3 -1] -3 5 3 6 7 1
1 [3 -1 -3] 5 3 6 7 -1
1 3 [-1 -3 5] 3 6 7 -1
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] 6
Therefore, return the median sliding window as [1,-1,-1,3,5,6].

Note:
You may assume k is always valid, ie: k is always smaller than input array’s size for non-empty array.

题意很简单，和之前的这一道题leetcode 295. Find Median from Data Stream 双优先级队列 + 中位数查找 思路一样，就这么做吧

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>using namespace std;class Solution {public:    vector<double> medianSlidingWindow(vector<int>& nums, int k)    {        vector<double> medians;        map<int, int> hash;                          // count numbers to be deleted        priority_queue<int, vector<int>> bheap;                // heap on the bottom        priority_queue<int, vector<int>, greater<int>> theap;  // heap on the top        int i = 0;        // Initialize the heaps        while (i < k)         {             bheap.push(nums[i++]);         }        for (int count = k / 2; count > 0; --count)        {            theap.push(bheap.top());             bheap.pop();        }        while (true)        {            // Get median            if (k % 2)                 medians.push_back(bheap.top());            else                 medians.push_back(((double)bheap.top() + theap.top()) / 2);            if (i == nums.size())                 break;            int m = nums[i - k], n = nums[i++], balance = 0;            // What happens to the number m that is moving out of the window            if (m <= bheap.top())             {                 --balance;                  if (m == bheap.top())                    bheap.pop();                 else                     ++hash[m];            }            else            {                 ++balance;                  if (m == theap.top())                     theap.pop();                 else                     ++hash[m];             }            // Insert the new number n that enters the window            if (!bheap.empty() && n <= bheap.top())             {                 ++balance;                 bheap.push(n);             }            else            {                 --balance;                theap.push(n);            }            // Rebalance the bottom and top heaps            if (balance < 0)             {                 bheap.push(theap.top());                theap.pop();            }            else if (balance > 0)             {                 theap.push(bheap.top());                 bheap.pop();             }            // Remove numbers that should be discarded at the top of the two heaps            while (!bheap.empty() && hash[bheap.top()])            {                --hash[bheap.top()];                bheap.pop();             }            while (!theap.empty() && hash[theap.top()])             {                 --hash[theap.top()];                theap.pop();            }        }        return medians;    }};