leetcode200-dfs-岛屿的数量

/**
<p>给你一个由 <code>'1'</code>（陆地）和 <code>'0'</code>（水）组成的的二维网格，请你计算网格中岛屿的数量。</p>

<p>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。</p>

<p>此外，你可以假设该网格的四条边均被水包围。</p>

<p> </p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
<strong>输出：</strong>1
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
<strong>输出：</strong>3
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 <= m, n <= 300</code></li>
	<li><code>grid[i][j]</code> 的值为 <code>'0'</code> 或 <code>'1'</code></li>
</ul>
<div><div>Related Topics</div><div><li>深度优先搜索</li><li>广度优先搜索</li><li>并查集</li><li>数组</li><li>矩阵</li></div></div><br><div><li>👍 1770</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {

	//主函数计算岛屿数量
	public int numIslands(char[][] grid) {
		int res = 0;
		int m = grid.length;
		int n = grid[0].length;
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if(grid[i][j]=='1'){
					res++;
					dfs(grid,i,j);
				}
			}
		}
		return res;
	}

	//子函数负责淹没 flood fill,
	void dfs(char[][] grid,int i ,int j){
		int m = grid.length;
		int n = grid[0].length;
		if (i < 0 || j < 0 || i >= m || j >= n) {
			return ;
		}

		if(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);
	}
}
//leetcode submit region end(Prohibit modification and deletion)