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

心已赠人 2022-06-03 02:55 124阅读 0赞

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][leetcode 295. Find Median from Data Stream _ _] 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;
        }
    };

也可以这么做

**十分需要注意的是，这里使用multiset是为了考虑重复元素，注意删除元素的时刻不可以直接使用erase(key)，因为这样会删除所有的key，应该这么做erase(find(key))才会删除一个元素key**

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    
    using namespace std;
    
    class Solution 
    {
    public:
        vector<double> medianSlidingWindow(vector<int>& nums, int k) 
        {
            vector<double> res;
            multiset<long long> st;
            for (int i = 0; i < k - 1; i++)
                st.insert((long long)nums[i]);
            for (int i = k - 1; i < nums.size(); i++)
            {
                st.insert(nums[i]);
                vector<long long> tmp(st.begin(), st.end());
                if (k % 2 == 0)
                    res.push_back((tmp[k / 2] + tmp[k / 2 - 1]) / 2.0);
                else
                    res.push_back(tmp[k/2]);
    
                st.erase(st.find(nums[i - k + 1]));
            }
            return res;
        }
    };

[leetcode 295. Find Median from Data Stream _ _]: http://blog.csdn.net/jackzhang_123/article/details/78121691