岛屿数量(深度优先遍历)

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'

• class Solution {
      int m, n, count = 0;
      ;
      boolean[][] vis;
      int []dx = {0, 0, 1, -1};
      int[] dy = {1, -1, 0, 0};
  
      public int numIslands(char[][] array) {
          m = array.length;
          n = array[0].length;
          vis = new boolean[m][n];
          for (int i = 0; i < m; i++) {
              for (int j = 0; j < n; j++) {
                  if (array[i][j] == '1' && !vis[i][j]) {
                      dfs(array, i, j);
                      count++;
                  }
              }
          }
          return count;
      }
      private void dfs(char [][]array,int i,int j){
          vis[i][j]=true;//标记陆地
          for(int k=0;k<4;k++){
              int x=i+dx[k],y=j+dy[k];
              if(x>=0&&x<m&&y>=0&&y<n&&!vis[x][y]){
                  if(array[x][y]=='1')dfs(array,x,y);
              }
          }
      }
  }