LeetCode：200. 岛屿数量

问题链接

200. 岛屿数量

问题描述

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

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

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

解题思路

采用沉岛策略，遍历网格，当发现岛屿（找到“1”）时，将这整一块岛屿沉入海底（利用深度优先搜索算法）。

代码示例（JAVA）

class Solution {
    int[] arrX = {1, -1, 0, 0};
    int[] arrY = {0, 0, -1, 1};

    public int numIslands(char[][] grid) {
        int count = 0;
        int len = grid.length;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    grid[i][j] = '0';
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    public void dfs(char[][] grid, int x, int y) {
        for (int i = 0; i < arrX.length; i++) {
            if (x + arrX[i] >= 0 && x + arrX[i] < grid.length && y + arrY[i] >= 0 && y + arrY[i] < grid[0].length
                    && grid[x + arrX[i]][y + arrY[i]] == '1') {
                grid[x + arrX[i]][y + arrY[i]] = '0';
                dfs(grid, x + arrX[i], y + arrY[i]);
            }
        }
    }
}