200:岛屿数量

src/main/java/leetcode/editor/cn/NumberOfIslands.java

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+//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' 
+// 
+// Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵 
+// 👍 1212 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.Arrays;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
+
+//200:岛屿数量
+public class NumberOfIslands {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new NumberOfIslands().new Solution();
+        System.out.println(solution.numIslands(new char[][]{
+                {'1', '1', '1', '1', '0'},
+                {'1', '1', '0', '1', '0'},
+                {'1', '1', '0', '0', '0'},
+                {'0', '0', '0', '0', '0'}
+        }));
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int numIslands(char[][] grid) {
+            int count = 0;
+            for(int i = 0; i < grid.length; i++) {
+                for(int j = 0; j < grid[0].length; j++) {
+                    if(grid[i][j] == '1'){
+                        dfs(grid, i, j);
+                        count++;
+                    }
+                }
+            }
+            return count;
+        }
+        private void dfs(char[][] grid, int x, int y){
+            if(x < 0 || y < 0 || x >= grid.length || y >= grid[0].length || grid[x][y] == '0') {
+                return;
+            }
+            grid[x][y] = '0';
+            dfs(grid, x + 1, y);
+            dfs(grid, x, y + 1);
+            dfs(grid, x - 1, y);
+            dfs(grid, x, y - 1);
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/NumberOfIslands.md

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+<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>\n<div><li>👍 1212</li><li>👎 0</li></div>