200. 岛屿数量(中等)

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1，问题描述

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：

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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：

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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'

2，初步思考

​ 解法1：染色标记

​ 解法2：hash缓存

3，代码处理

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import java.util.ArrayList;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

public class _200岛屿数量 {

    // 解法：染色标记
    public int numIslands_color(char[][] grid) {
        rows = grid.length;
        columns = grid[0].length;// 前行后列
        cnt = 0;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                dfs(grid, i, j, true);
            }
        }
        return cnt;
    }

    private void dfs(char[][] grid, int i, int j, boolean isStart) {
        if (i < 0 || i >= rows || j < 0 || j >= columns || grid[i][j] != '1') {
            return;
        }
        if (isStart) {
            cnt++;
        }
        grid[i][j] = '2';
        dfs(grid, i - 1, j, false);
        dfs(grid, i + 1, j, false);
        dfs(grid, i, j - 1, false);
        dfs(grid, i, j + 1, false);
    }

    // 字典处理
    public int numIslands(char[][] grid) {
        Set<String> set = new HashSet<>();
        rows = grid.length;
        columns = grid[0].length;// 前行后列
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (grid[i][j] == '1') {
                    set.add(i + "," + j);
                }
            }
        }

        int cnt = 0;
        while (!set.isEmpty()) {
            String key = new ArrayList<>(set).getFirst();
            dfs(grid, set, key);
            cnt++;
        }
        return cnt;
    }

    int rows, columns, cnt;

    // 递归处理
    private void dfs(char[][] grid, Set<String> set, String key) {
        set.remove(key);
        String[] split = key.split(",");
        int i = Integer.parseInt(split[0]), j = Integer.parseInt(split[1]);
        String key1 = (i - 1) + "," + j;// 上
        String key2 = (i + 1) + "," + j;// 下
        String key3 = i + "," + (j - 1);// 左
        String key4 = i + "," + (j + 1);// 右
        if (set.contains(key1)) {
            dfs(grid, set, key1);
        }
        if (set.contains(key2)) {
            dfs(grid, set, key2);
        }
        if (set.contains(key3)) {
            dfs(grid, set, key3);
        }
        if (set.contains(key4)) {
            dfs(grid, set, key4);
        }
    }

    public static void main(String[] args) {
        _200岛屿数量 island = new _200岛屿数量();
//        System.out.println(island.numIslands(new char[][]{
//                {'1', '1', '1', '1', '0'},
//                {'1', '1', '0', '1', '0'},
//                {'1', '1', '0', '0', '0'},
//                {'0', '0', '0', '0', '0'}
//        }));
//        System.out.println(island.numIslands(new char[][]{
//                {'1', '1', '0', '0', '0'},
//                {'1', '1', '0', '0', '0'},
//                {'0', '0', '1', '0', '0'},
//                {'0', '0', '0', '1', '1'}
//        }));
        System.out.println(island.numIslands(new char[][]{
                {'1', '1', '1'},
                {'1', '0', '1'},
                {'1', '1', '1'}
        }));
    }
}