01岛屿数量

岛屿数量

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'

void dfs(char[][] grid, int r, int c) {
  int nr = grid.length;
  int nc = grid[0].length;

  if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
    return;
  }

  grid[r][c] = '0';
  dfs(grid, r - 1, c);
  dfs(grid, r + 1, c);
  dfs(grid, r, c - 1);
  dfs(grid, r, c + 1);
}

public int numIslands(char[][] grid) {
  if (grid == null || grid.length == 0) {
    return 0;
  }

  int nr = grid.length;
  int nc = grid[0].length;
  int num_islands = 0;
  for (int r = 0; r < nr; ++r) {
    for (int c = 0; c < nc; ++c) {
      if (grid[r][c] == '1') {
        ++num_islands;
        dfs(grid, r, c);
      }
    }
  }

  return num_islands;
}

public int numIslands(char[][] grid) {
  int m = grid.length;
  int n = grid[0].length;

  int lands = 0;
  Queue<Integer> queue = new LinkedList<>();

  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if(grid[i][j]=='1'){
        lands++;
        queue.offer(i*n+j);
        while (!queue.isEmpty()){
          Integer poll = queue.poll();
          int i1 = poll/n;
          int j1 = poll%n;
          grid[i1][j1]='0';
          if(i1-1>=0 && grid[i1-1][j1]=='1'){
            queue.offer((i1-1)*n+j1);
          }
          if(i1+1<m && grid[i1+1][j1]=='1'){
            queue.offer((i1+1)*n+j1);
          }
          if(j1-1>=0 && grid[i1][j1-1]=='1'){
            queue.offer(i1*n+j1-1);
          }
          if(j1+1<n && grid[i1][j1+1]=='1'){
            queue.offer(i1*n+j1+1);
          }
        }
      }
    }
  }
  return lands;
}