diff --git a/problems/0309.最佳买卖股票时机含冷冻期.md b/problems/0309.最佳买卖股票时机含冷冻期.md
index 5a38111a..95689a48 100644
--- a/problems/0309.最佳买卖股票时机含冷冻期.md
+++ b/problems/0309.最佳买卖股票时机含冷冻期.md
@@ -359,6 +359,26 @@ func max(a, b int) int {
 
 ### Javascript:
 
+> 不同的状态定义 感觉更容易理解些
+```javascript
+function maxProfit(prices) {
+  // 第i天状态 持股 卖出 非冷冻期(不持股) 处于冷冻期
+  const dp = new Array(prices.length).fill(0).map(() => [0, 0, 0, 0]);
+  dp[0][0] = -prices[0];
+  for (let i = 1; i < prices.length; i++) {
+    // 持股
+    dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][2] - prices[i]);
+    // 卖出
+    dp[i][1] = dp[i - 1][0] + prices[i];
+    // 非冷冻期（不持股）
+    dp[i][2] = Math.max(dp[i - 1][2], dp[i - 1][1]);
+    // 冷冻期（上一天卖出）
+    dp[i][3] = dp[i - 1][1];
+  }
+  return Math.max(...dp.pop());
+};
+```
+
 ```javascript
 const maxProfit = (prices) => {
     if(prices.length < 2) {
diff --git a/problems/kamacoder/0099.岛屿的数量广搜.md b/problems/kamacoder/0099.岛屿的数量广搜.md
index b9bdb796..3c069b44 100644
--- a/problems/kamacoder/0099.岛屿的数量广搜.md
+++ b/problems/kamacoder/0099.岛屿的数量广搜.md
@@ -190,10 +190,164 @@ int main() {
 
 ### Java 
 
+```java
+
+import java.util.Scanner;
+
+public class Main {
+    static int[][] dir = { {0, 1}, {1, 0}, {-1, 0}, {0, -1} }; // 四个方向
+
+    public static void dfs(int[][] grid, boolean[][] visited, int x, int y) {
+        for (int i = 0; i < 4; i++) {
+            int nextx = x + dir[i][0];
+            int nexty = y + dir[i][1];
+            if (nextx < 0 || nextx >= grid.length || nexty < 0 || nexty >= grid[0].length) continue;  // 越界了，直接跳过
+            if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 没有访问过的 同时 是陆地的
+                visited[nextx][nexty] = true;
+                dfs(grid, visited, nextx, nexty);
+            }
+        }
+    }
+
+    public static void main(String[] args) {
+        Scanner scanner = new Scanner(System.in);
+        int n = scanner.nextInt();
+        int m = scanner.nextInt();
+        int[][] grid = new int[n][m];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                grid[i][j] = scanner.nextInt();
+            }
+        }
+
+        boolean[][] visited = new boolean[n][m];
+
+        int result = 0;
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                if (!visited[i][j] && grid[i][j] == 1) {
+                    visited[i][j] = true;
+                    result++; // 遇到没访问过的陆地，+1
+                    dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
+                }
+            }
+        }
+
+        System.out.println(result);
+    }
+}
+
+
+
+```
+
+
 ### Python
 
+```python
+
+def dfs(grid, visited, x, y):
+    dir = [(0, 1), (1, 0), (-1, 0), (0, -1)]  # 四个方向
+    for d in dir:
+        nextx, nexty = x + d[0], y + d[1]
+        if 0 <= nextx < len(grid) and 0 <= nexty < len(grid[0]):
+            if not visited[nextx][nexty] and grid[nextx][nexty] == 1:  # 没有访问过的 同时 是陆地的
+                visited[nextx][nexty] = True
+                dfs(grid, visited, nextx, nexty)
+
+def main():
+    n, m = map(int, input().split())
+    grid = [list(map(int, input().split())) for _ in range(n)]
+    visited = [[False] * m for _ in range(n)]
+
+    result = 0
+    for i in range(n):
+        for j in range(m):
+            if not visited[i][j] and grid[i][j] == 1:
+                visited[i][j] = True
+                result += 1  # 遇到没访问过的陆地，+1
+                dfs(grid, visited, i, j)  # 将与其链接的陆地都标记上 True
+
+    print(result)
+
+if __name__ == "__main__":
+    main()
+
+
+
+```
+
+
 ### Go
 
+```go
+
+package main
+
+import (
+    "bufio"
+    "fmt"
+    "os"
+    "strconv"
+    "strings"
+)
+
+var dir = [4][2]int{{0, 1}, {1, 0}, {-1, 0}, {0, -1}} // 四个方向
+
+func dfs(grid [][]int, visited [][]bool, x, y int) {
+    for i := 0; i < 4; i++ {
+        nextx := x + dir[i][0]
+        nexty := y + dir[i][1]
+        if nextx < 0 || nextx >= len(grid) || nexty < 0 || nexty >= len(grid[0]) {
+            continue // 越界了，直接跳过
+        }
+        if !visited[nextx][nexty] && grid[nextx][nexty] == 1 { // 没有访问过的 同时 是陆地的
+            visited[nextx][nexty] = true
+            dfs(grid, visited, nextx, nexty)
+        }
+    }
+}
+
+func main() {
+    reader := bufio.NewReader(os.Stdin)
+    var n, m int
+    fmt.Scanf("%d %d", &n, &m)
+
+    grid := make([][]int, n)
+    for i := 0; i < n; i++ {
+        grid[i] = make([]int, m)
+        line, _ := reader.ReadString('\n')
+        line = strings.TrimSpace(line)
+        elements := strings.Split(line, " ")
+        for j := 0; j < m; j++ {
+            grid[i][j], _ = strconv.Atoi(elements[j])
+        }
+    }
+
+    visited := make([][]bool, n)
+    for i := 0; i < n; i++ {
+        visited[i] = make([]bool, m)
+    }
+
+    result := 0
+    for i := 0; i < n; i++ {
+        for j := 0; j < m; j++ {
+            if !visited[i][j] && grid[i][j] == 1 {
+                visited[i][j] = true
+                result++ // 遇到没访问过的陆地，+1
+                dfs(grid, visited, i, j) // 将与其链接的陆地都标记上 true
+            }
+        }
+    }
+
+    fmt.Println(result)
+}
+
+
+```
+
+
+
 ### Rust
 
 ### Javascript
@@ -283,12 +437,62 @@ const bfs = (graph, visited, x, y) => {
 
 ### PhP
 
+```PHP
+
+<?php
+
+function dfs($grid, &$visited, $x, $y) {
+    $dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
+    foreach ($dir as $d) {
+        $nextx = $x + $d[0];
+        $nexty = $y + $d[1];
+        if ($nextx < 0 || $nextx >= count($grid) || $nexty < 0 || $nexty >= count($grid[0])) {
+            continue; // 越界了，直接跳过
+        }
+        if (!$visited[$nextx][$nexty] && $grid[$nextx][$nexty] == 1) { // 没有访问过的 同时 是陆地的
+            $visited[$nextx][$nexty] = true;
+            dfs($grid, $visited, $nextx, $nexty);
+        }
+    }
+}
+
+function main() {
+    fscanf(STDIN, "%d %d", $n, $m);
+    $grid = [];
+    for ($i = 0; $i < $n; $i++) {
+        $grid[$i] = array_map('intval', explode(' ', trim(fgets(STDIN))));
+    }
+
+    $visited = array_fill(0, $n, array_fill(0, $m, false));
+
+    $result = 0;
+    for ($i = 0; $i < $n; $i++) {
+        for ($j = 0; $j < $m; $j++) {
+            if (!$visited[$i][$j] && $grid[$i][$j] == 1) {
+                $visited[$i][$j] = true;
+                $result++; // 遇到没访问过的陆地，+1
+                dfs($grid, $visited, $i, $j); // 将与其链接的陆地都标记上 true
+            }
+        }
+    }
+
+    echo $result . PHP_EOL;
+}
+
+main();
+?>
+
+
+```
+
+
 ### Swift
 
 ### Scala
 
 ### C#
 
+
 ### Dart
 
 ### C
diff --git a/problems/kamacoder/0100.岛屿的最大面积.md b/problems/kamacoder/0100.岛屿的最大面积.md
index 43e7e968..024e509f 100644
--- a/problems/kamacoder/0100.岛屿的最大面积.md
+++ b/problems/kamacoder/0100.岛屿的最大面积.md
@@ -226,6 +226,100 @@ public:
 
 ### Python
 
+DFS
+
+```python
+# 四个方向
+position = [[0, 1], [1, 0], [0, -1], [-1, 0]]
+count = 0
+
+
+def dfs(grid, visited, x, y):
+    """
+    深度优先搜索，对一整块陆地进行标记
+    """
+    global count  # 定义全局变量，便于传递count值
+    for i, j in position:
+        cur_x = x + i
+        cur_y = y + j
+        # 下标越界，跳过
+        if cur_x < 0 or cur_x >= len(grid) or cur_y < 0 or cur_y >= len(grid[0]):
+            continue
+        if not visited[cur_x][cur_y] and grid[cur_x][cur_y] == 1:
+            visited[cur_x][cur_y] = True
+            count += 1
+            dfs(grid, visited, cur_x, cur_y)
+
+
+n, m = map(int, input().split())
+# 邻接矩阵
+grid = []
+for i in range(n):
+    grid.append(list(map(int, input().split())))
+# 访问表
+visited = [[False] * m for _ in range(n)]
+
+result = 0  # 记录最终结果
+for i in range(n):
+    for j in range(m):
+        if grid[i][j] == 1 and not visited[i][j]:
+            count = 1
+            visited[i][j] = True
+            dfs(grid, visited, i, j)
+            result = max(count, result)
+
+print(result)
+```
+
+BFS
+
+```python
+from collections import deque
+
+position = [[0, 1], [1, 0], [0, -1], [-1, 0]]  # 四个方向
+count = 0
+
+
+def bfs(grid, visited, x, y):
+    """
+    广度优先搜索对陆地进行标记
+    """
+    global count  # 声明全局变量
+    que = deque()
+    que.append([x, y])
+    while que:
+        cur_x, cur_y = que.popleft()
+        for i, j in position:
+            next_x = cur_x + i
+            next_y = cur_y + j
+            # 下标越界，跳过
+            if next_x < 0 or next_x >= len(grid) or next_y < 0 or next_y >= len(grid[0]):
+                continue
+            if grid[next_x][next_y] == 1 and not visited[next_x][next_y]:
+                visited[next_x][next_y] = True
+                count += 1
+                que.append([next_x, next_y])
+
+
+n, m = map(int, input().split())
+# 邻接矩阵
+grid = []
+for i in range(n):
+    grid.append(list(map(int, input().split())))
+visited = [[False] * m for _ in range(n)]  # 访问表
+
+result = 0  # 记录最终结果
+for i in range(n):
+    for j in range(m):
+        if grid[i][j] == 1 and not visited[i][j]:
+            count = 1
+            visited[i][j] = True
+            bfs(grid, visited, i, j)
+            res = max(result, count)
+
+print(result)
+```
+
 ### Go
 
 ### Rust