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+# [661. Image Smoother](https://leetcode.com/problems/image-smoother/description/)
+# 思路
+将二维数组遍历一遍，每次按照题意计算avg值，实现相邻的小trick:   
+```
+  int dx[8] = {1, -1, 1, -1, 1, 0, -1, 0};
+  int dy[8] = {-1, 1, 1, -1, 0, 1, 0, -1};
+  for(int k = 0; k < 8; k++){
+    new_x = x + dx[k];
+    new_y = y + dy[k];
+  }
+```
+另外要注意判断以上代码算出来的相邻坐标是否合法。   
+空间上改进：   
+由于元素值为0-255， 所以最多8位，所以可以先用int型的高8位存放新的值，这样就可以使空间复杂度为O(1)。
+
+# C++
+```
+class Solution {
+public:
+    bool islegal(const int& x, const int& y, const int& r, const int &c){
+        if(x < 0 || y < 0 || x >= r || y >= c) return false;
+        return true;
+    }
+    vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
+        int dx[8] = {1, -1, 1, -1, 1, 0, -1, 0};
+        int dy[8] = {-1, 1, 1, -1, 0, 1, 0, -1};
+        
+        int r = M.size(), c = M[0].size(), sum, count;
+        vector<vector<int>>res;
+        vector<int>tmp;
+        for(int i = 0; i < r; i++){
+            for(int j = 0; j < c; j++){
+                sum = M[i][j], count = 1;
+                for(int k = 0; k < 8; k++){
+                    if(islegal(i + dx[k], j + dy[k], r, c)) {
+                        sum += M[i + dx[k]][j + dy[k]];
+                        count++;
+                    }
+                }
+                tmp.push_back(sum / count);      
+            }
+            res.push_back(tmp);
+            tmp.clear();
+        }
+        return res;
+    }
+};
+```