岛屿问题(dfs)

leetcode 200 岛屿数量

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

  private void dfs(char[][] grid, int i, int j) {
    if(i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0'){
      return;
    }
    grid[i][j] = '0';
    dfs(grid,i+1,j);
    dfs(grid,i,j+1);
    dfs(grid,i-1,j);
    dfs(grid,i,j-1);
  }
}

leetcode 695 岛屿的最大数量

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        if (grid == null) {
            return 0;
        }
        int res = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                int area = dfs(grid, i, j);
                res = Math.max(area, res);
            }
        }
        return res;
    }

    private int dfs(int[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == 0) {
            return 0;
        }
        grid[i][j] = 0;
        return 1 + dfs(grid, i + 1, j) 
        + dfs(grid, i, j + 1) 
        + + dfs(grid, i - 1, j) 
        + + dfs(grid, i, j - 1);
    }
}

leetcode 463 岛屿的周长

class Solution {
    public int islandPerimeter(int[][] grid) {
        if (grid == null) {
            return 0;
        }
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    return dfs(grid, i, j);
                }
            }
        }
        return 0;
    }

    private int dfs(int[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) {
            return 1;
        }
        if (grid[i][j] == 0) {
            return 1;// 对应海洋
        }
        if (grid[i][j] != 1) {// 当前是已经遍历过的陆地格子
            return 0;
        }
        grid[i][j] = 2;
        return dfs(grid, i + 1, j) + dfs(grid, i, j + 1) + dfs(grid, i - 1, j) + dfs(grid, i, j - 1);
    }
}