leetcode695-dfs

/**
<p>给你一个大小为 <code>m x n</code> 的二进制矩阵 <code>grid</code> 。</p>

<p><strong>岛屿</strong>&nbsp;是由一些相邻的&nbsp;<code>1</code>&nbsp;(代表土地) 构成的组合，这里的「相邻」要求两个 <code>1</code> 必须在 <strong>水平或者竖直的四个方向上 </strong>相邻。你可以假设&nbsp;<code>grid</code> 的四个边缘都被 <code>0</code>（代表水）包围着。</p>

<p>岛屿的面积是岛上值为 <code>1</code> 的单元格的数目。</p>

<p>计算并返回 <code>grid</code> 中最大的岛屿面积。如果没有岛屿，则返回面积为 <code>0</code> 。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/01/maxarea1-grid.jpg" style="width: 500px; height: 310px;" />
<pre>
<strong>输入：</strong>grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
<strong>输出：</strong>6
<strong>解释：</strong>答案不应该是 <code>11</code> ，因为岛屿只能包含水平或垂直这四个方向上的 <code>1</code> 。
</pre>

<p><strong>示例 2：</strong></p>

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

<p>&nbsp;</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 &lt;= m, n &lt;= 50</code></li>
	<li><code>grid[i][j]</code> 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>👍 800</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
	
	//淹没岛屿的同时计算面积
	public int maxAreaOfIsland(int[][] grid) {
		int m = grid.length;
		int n = grid[0].length;

		int res = 0;
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if(grid[i][j]==1){
					res = Math.max(res,dfs(grid,i,j));
				}
			}
		}
		return res;
	}

	int dfs(int[][] grid, int i, int j) {
		int m = grid.length;
		int n = grid[0].length;

		if (i < 0 || j < 0 || i >= m || j >= n) {
			return 0;
		}

		if (grid[i][j] == 0) {
			return 0;
		}
		grid[i][j] = 0;

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

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