package LeetCode;

public class LeetCode200 {

//https://leetcode-cn.com/problems/number-of-islands/description/

/// 时间复杂度: O(n*m)

/// 空间复杂度: O(n*m)

private int d[][] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

private int m, n;

private boolean visited[][];

public int numIslands(char[][] grid) {

if (grid == null || grid.length == 0 || grid[0].length == 0)

return 0;

m = grid.length;

n = grid[0].length;

visited = new boolean[m][n];

int res = 0;

for (int i = 0; i < m; i++)

for (int j = 0; j < n; j++)

if (grid[i][j] == '1' && !visited[i][j]) {

dfs(grid, i, j);

res++;

}

return res;

}

// 从grid[x][y]的位置开始,进行floodfill

// 保证(x,y)合法,且grid[x][y]是没有被访问过的陆地

private void dfs(char[][] grid, int x, int y) {

//assert(inArea(x,y));

visited[x][y] = true;

for (int i = 0; i < 4; i++) {

int newx = x + d[i][0];

int newy = y + d[i][1];

if (inArea(newx, newy) && !visited[newx][newy] && grid[newx][newy] == '1')

dfs(grid, newx, newy);

}

return;

}

private boolean inArea(int x, int y) {

return x >= 0 && x < m && y >= 0 && y < n;

}

public static void main(String[] args) {

char grid1[][] = {

{'1', '1', '1', '1', '0'},

{'1', '1', '0', '1', '0'},

{'1', '1', '0', '0', '0'},

{'0', '0', '0', '0', '0'}

};

System.out.println((new LeetCode200()).numIslands(grid1));

// 1

// ---

char grid2[][] = {

{'1', '1', '0', '0', '0'},

{'1', '1', '0', '0', '0'},

{'0', '0', '1', '0', '0'},

{'0', '0', '0', '1', '1'}

};

System.out.println((new LeetCode200()).numIslands(grid2));

// 3

}

}