技術標籤：刷題

難度：困難。
感覺沒有昨天的難呀。這個有點思路。
用一個大小為k的陣列有序的儲存每個視窗的數，每次視窗移動更新一次。
使用二分法刪除和插入數字。

正確解法：

class Solution {

    void update_vector(vector<int>& nums, int del_num, int add_num){
        int n = nums.size();
        int left = 0, right = n - 1, mid = 0;
        while(left <= right){
            mid =
 
 (left + right) / 2;
            if(nums[mid] < del_num){
                left = mid + 1;
            }
            else if(nums[mid] > del_num){
                right = mid - 1;
            }
            else{
                nums.erase(nums.begin() + mid);
                break;
            }
 

        }
        left = 0, right = n - 2;
        
        while(left <= right){
            mid = (left + right) / 2;
            if(nums[mid] < add_num){
                left = mid + 1;
            }
            else if(nums[mid] > add_num){
                right = mid - 1;
            }
            else
 
{
                right = mid;
                break;
            }
        }
        nums.insert(nums.begin() + right + 1, add_num);
    }

public:
    vector<double> medianSlidingWindow(vector<int>& nums, int k) {
        vector<double> ans;
        int n = nums.size();
        if(k == 1){
            for(int i = 0; i < n; i++){
                ans.push_back(nums[i]);
            }
            return ans;
        }
        
        vector<int> orderd_array(k);
        for(int i = 0; i < k; i++){
            orderd_array[i] = nums[i];
        }
        sort(orderd_array.begin(), orderd_array.end());

        int left = 0, right = k - 1;
        
        double temp1 = 0, temp2 = 0;
        if(k % 2){
            ans.push_back(orderd_array[k / 2]);
        }
        else{
            temp1 = orderd_array[k / 2 - 1];
            temp2 = orderd_array[k / 2];
            ans.push_back((temp1 + temp2) / 2.0);
        }

        while(right < n - 1){
            left++;
            right++;
            update_vector(orderd_array, nums[left - 1], nums[right]);
            if(k % 2){
                ans.push_back(orderd_array[k / 2]);
            }
            else{
                temp1 = orderd_array[k / 2 - 1];
                temp2 = orderd_array[k / 2];
                ans.push_back((temp1 + temp2) / 2.0);
            }
        }

        return ans;
    }
};