题目描述

给定一个由 '1'（表示陆地）和 '0'（表示水域）组成的二维网格，计算网格中岛屿的数量。岛屿总是被水域包围，并且每块岛屿由水平方向和垂直方向上相邻的陆地组成。此外，你可以假设该网格的四条边均被水域包围。

示例

输入:

11110
11010
11000
00000

输出: 1

解释: 只有一个岛屿，由左上角的四个 '1' 组成。

PHP 示例代码

function numIslands($grid) {
    if (empty($grid) || empty($grid[0])) {
        return 0;
    }

    $rows = count($grid);
    $cols = strlen($grid[0]);
    $count = 0;

    for ($i = 0; $i < $rows; $i++) {
        for ($j = 0; $j < $cols; $j++) {
            if ($grid[$i][$j] == '1') {
                dfs($grid, $i, $j);
                $count++;
            }
        }
    }

    return $count;
}

function dfs(&$grid, $i, $j) {
    $rows = count($grid);
    $cols = strlen($grid[0]);

    if ($i < 0 || $j < 0 || $i >= $rows || $j >= $cols || $grid[$i][$j] != '1') {
        return;
    }

    $grid[$i][$j] = '#'; // 标记为已访问

    dfs($grid, $i - 1, $j); // 上
    dfs($grid, $i + 1, $j); // 下
    dfs($grid, $i, $j - 1); // 左
    dfs($grid, $i, $j + 1); // 右
}

// 示例用法
$grid = [
    "11110",
    "11010",
    "11000",
    "00000"
];
echo numIslands($grid); // 输出: 1

Python 示例代码

def numIslands(grid):
    if not grid or not grid[0]:
        return 0

    rows, cols = len(grid), len(grid[0])
    count = 0

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == '1':
                dfs(grid, i, j)
                count += 1

    return count

def dfs(grid, i, j):
    rows, cols = len(grid), len(grid[0])

    if i < 0 or j < 0 or i >= rows or j >= cols or grid[i][j] != '1':
        return

    grid[i][j] = '#' # 标记为已访问

    dfs(grid, i - 1, j) # 上
    dfs(grid, i + 1, j) # 下
    dfs(grid, i, j - 1) # 左
    dfs(grid, i, j + 1) # 右

# 示例用法
grid = [
    ["1","1","1","1","0"],
    ["1","1","0","1","0"],
    ["1","1","0","0","0"],
    ["0","0","0","0","0"]
]
print(numIslands(grid)) # 输出: 1

JavaScript 示例代码

function numIslands(grid) {
    if (!grid.length || !grid[0].length) {
        return 0;
    }

    let rows = grid.length;
    let cols = grid[0].length;
    let count = 0;

    for (let i = 0; i < rows; i++) {
        for (let j = 0; j < cols; j++) {
            if (grid[i][j] === '1') {
                dfs(grid, i, j);
                count++;
            }
        }
    }

    return count;
}

function dfs(grid, i, j) {
    const rows = grid.length;
    const cols = grid[0].length;

    if (i < 0 || j < 0 || i >= rows || j >= cols || grid[i][j] !== '1') {
        return;
    }

    grid[i][j] = '#'; // 标记为已访问

    dfs(grid, i - 1, j); // 上
    dfs(grid, i + 1, j); // 下
    dfs(grid, i, j - 1); // 左
    dfs(grid, i, j + 1); // 右
}

// 示例用法
const grid = [
    ["1","1","1","1","0"],
    ["1","1","0","1","0"],
    ["1","1","0","0","0"],
    ["0","0","0","0","0"]
];
console.log(numIslands(grid)); // 输出: 1

这些示例代码均使用了深度优先搜索（DFS）算法来遍历并标记岛屿中的每个陆地，从而计算出岛屿的总数。