200. 岛屿数量

思路

深度优先遍历，两层for用于找到每个节点，dfs用于找到连续的一片岛并且把所有点置位0防止重复遍历

代码:

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


    /**
     * 搜索周边的1,并把四周为1的置为0;
     * @author zyh
     * @date 2021/11/3
     */
    private void dfs(char[][] grid, int i, int j) {
        if (i<0||i>=grid.length||j<0||j>=grid[0].length||grid[i][j]=='0'){
            return;
        }
        grid[i][j]='0';

        dfs(grid,i-1,j);
        dfs(grid,i+1,j);
        dfs(grid,i,j-1);
        dfs(grid,i,j+1);
    }


}