Leetcode 480. Sliding Window Median

发布时间：2020-12-14 02:33:34 所属栏目：Windows 来源：网络整理

导读：Problem: ? 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)

Problem:

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,-1,-3,5,6,7],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 [1,6].

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

Solution:

　　这道题思路不难，但写起来有点麻烦，看到这道题，发现和Leetcode 239. Sliding Window Maximum非常相似，但最大值和中值的处理却几乎完全不同，然后考虑用双优先级队列，但优先级队列的删除操作却非常影响效率，既要删除比较快，又要速度快，map是最好的选择了，一般来说，能用优先级队列的题目用map也能解决。考虑到可能存在重复的数字，所以用left和right两个变量记录small和big两个map的数字和（不能简单的用map.size()），动态添加和删除即可。

Code:

 1 class Solution {
 2 public:
 3     void addValue(map<int,int> &small,map<int,int> &big,int &left,int &right,int &value){
 4         if(left == right){
 5             if(small.empty() || value <= small.rbegin()->first){
 6                 small[value]++;
 7             }
 8             else{
 9                 big[value]++;
10                 small[big.begin()->first]++;
11                 big[big.begin()->first]--;
12                 if(big[big.begin()->first] == 0)
13                     big.erase(big.begin()->first);
14             }
15             left++;
16         }
17         else{
18             if(value <= small.rbegin()->first){
19                 small[value]++;
20                 big[small.rbegin()->first]++;
21                 small[small.rbegin()->first]--;
22                 if(small[small.rbegin()->first] == 0)
23                     small.erase(small.rbegin()->first);
24             }
25             else{
26                 big[value]++;
27             }
28             right++;
29         }
30     }
31     vector<double> medianSlidingWindow(vector<int>& nums,int k) {
32         vector<double> result;
33         bool flag = k%2;
34         int left = 0,right = 0;
35         map<int,int> small,big;
36         for(int i = 0;i != k;++i){
37             addValue(small,big,left,right,nums[i]);
38         }
39         result.push_back(flag?(small.rbegin()->first):((int64_t)(small.rbegin()->first)+(int64_t)(big.begin()->first))/2.0);
40         for(int i = k;i != nums.size();++i){
41             int prev = nums[i-k];
42             if(prev <= small.rbegin()->first){
43                 small[prev]--;
44                 if(small[prev] == 0)
45                     small.erase(prev);
46                 if(!flag){
47                     small[big.begin()->first]++;
48                     big[big.begin()->first]--;
49                     if(big[big.begin()->first] == 0)
50                         big.erase(big.begin()->first);
51                     right--;
52                 }
53                 else
54                     left--;
55             }
56             else{
57                 big[prev]--;
58                 if(big[prev] == 0)
59                     big.erase(prev);
60                 if(flag){
61                     big[small.rbegin()->first]++;
62                     small[small.rbegin()->first]--;
63                     if(small[small.rbegin()->first] == 0)
64                         small.erase(small.rbegin()->first);
65                     left--;
66                 }
67                 else
68                     right--;
69             }
70             addValue(small,nums[i]);
71             result.push_back(flag?(small.rbegin()->first):((int64_t)(small.rbegin()->first)+(int64_t)(big.begin()->first))/2.0);
72         }
73         return result;
74     }
75 };

（编辑：李大同）

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