diff --git a/problems/0463.岛屿的周长.md b/problems/0463.岛屿的周长.md
index ddd43d8c..b9d34b84 100644
--- a/problems/0463.岛屿的周长.md
+++ b/problems/0463.岛屿的周长.md
@@ -120,6 +120,41 @@ class Solution {
 ```
 
 Python：
+### 解法1:
+扫描每个cell,如果当前位置为岛屿 grid[i][j] == 1， 从当前位置判断四边方向，如果边界或者是水域，证明有边界存在，res矩阵的对应cell加一。
+
+```python3
+class Solution:
+    def islandPerimeter(self, grid: List[List[int]]) -> int:
+
+        m = len(grid)
+        n = len(grid[0])
+
+        # 创建res二维素组记录答案
+        res = [[0] * n for j in range(m)]
+
+        for i in range(m):
+            for j in range(len(grid[i])):
+                # 如果当前位置为水域，不做修改或reset res[i][j] = 0
+                if grid[i][j] == 0:
+                    res[i][j] = 0
+                # 如果当前位置为陆地，往四个方向判断，update res[i][j]
+                elif grid[i][j] == 1:
+                    if i == 0 or (i > 0 and grid[i-1][j] == 0):
+                        res[i][j] += 1
+                    if j == 0 or (j >0 and grid[i][j-1] == 0):
+                        res[i][j] += 1
+                    if i == m-1 or (i < m-1 and grid[i+1][j] == 0):
+                        res[i][j] += 1
+                    if j == n-1 or (j < n-1 and grid[i][j+1] == 0):
+                        res[i][j] += 1
+
+        # 最后求和res矩阵，这里其实不一定需要矩阵记录，可以设置一个variable res 记录边长，舍矩阵无非是更加形象而已
+        ans = sum([sum(row) for row in res])
+
+        return ans
+
+```
 
 Go：