/*

    思路是：遍历二维数组，遇到1就把周围连续的1变成0，res+1，然后继续遍历，直到结束

    周围连续1置零用的是dfs，向四个位置搜索，遇到0返回

     */

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

        if (grid.length==0) return 0;

        int r = grid.length;

        int c = grid[0].length;

        int res = 0;

        //遍历数组

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

            for (int j = 0; j < c; j++) {

                if (grid[i][j]=='1')

                {

                    res++;

                    //进行置零操作

                    dfs(grid,i,j,r,c);

                }

            }

        }

        return res;

    }

    public void dfs(char[][] grid,int i,int j,int r,int c)

    {

        //出界和遇0返回

        if (i<0||j<0||i>=r||j>=c||grid[i][j]!='1') return;

        //置零

        grid[i][j]='0';

        //把周围连续的置零

        dfs(grid,i-1,j,r,c);

        dfs(grid,i+1,j,r,c);

        dfs(grid,i,j-1,r,c);

        dfs(grid,i,j+1,r,c);

    }