Create 200. 岛屿数量.md

Author: CompetitiveLin

Depth First Search/200. 岛屿数量.md

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+#### 200. 岛屿数量
+
+难度：中等
+
+---
+
+给你一个由 `'1'`（陆地）和 `'0'`（水）组成的的二维网格，请你计算网格中岛屿的数量。
+
+岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
+
+此外，你可以假设该网格的四条边均被水包围。
+
+ **示例 1：** 
+
+```
+输入：grid = [
+  ["1","1","1","1","0"],
+  ["1","1","0","1","0"],
+  ["1","1","0","0","0"],
+  ["0","0","0","0","0"]
+]
+输出：1
+```
+
+ **示例 2：** 
+
+```
+输入：grid = [
+  ["1","1","0","0","0"],
+  ["1","1","0","0","0"],
+  ["0","0","1","0","0"],
+  ["0","0","0","1","1"]
+]
+输出：3
+```
+
+ **提示：** 
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   `1 <= m, n <= 300`
+*   `grid[i][j]` 的值为 `'0'` 或 `'1'`
+
+---
+
+深度搜索优先：
+
+遍历每一格数据，如果当前格数据为 `1`，则岛屿数量加一，并且将四周（上下左右）的 `1` 变为 `0`，不断扩散（深度搜索）
+
+```Go
+func numIslands(grid [][]byte) int {
+    res := 0
+    var dfs func(grid [][]byte, i, j int)
+    dfs = func(grid [][]byte, i, j int) {
+        if grid[i][j] == '0' {
+            return
+        }
+        m, n := len(grid), len(grid[0])
+        grid[i][j] = '0'
+        if i > 0 {
+            dfs(grid, i - 1, j)
+        }
+        if i < m - 1 {
+            dfs(grid, i + 1, j)
+        }
+        if j > 0 {
+            dfs(grid, i, j - 1)
+        }
+        if j < n - 1 {
+            dfs(grid, i, j + 1)
+        }
+    }
+    for i := 0; i < len(grid); i++ {
+        for j := 0; j < len(grid[0]); j++ {
+            if grid[i][j] == '1' {
+                dfs(grid, i, j)
+                res ++
+            }
+        }
+    }
+    return res
+}
+```