200. 岛屿数量 (opens new window)

English Version (opens new window)

题目描述

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'

解法

BFS/DFS 均可

Python3

Java

TypeScript

function numIslands(grid: string[][]): number {
    let m = grid.length, n = grid[0].length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == '1') {
                dfs(grid, i, j);
                ++ans;
            }
        }
    }
    return ans;
};

function dfs(grid: string[][], i: number, j: number) {
    let m = grid.length, n = grid[0].length;
    if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || grid[i][j] == '0') {
        return;
    }
    grid[i][j] = '0';
    for (let [dx, dy] of [[0, 1], [0, -1], [1, 0], [-1, 0]]) {
        let x = i + dx, y = j + dy;
        dfs(grid, x, y);
    }
}

...