feat: 被围绕的岛屿 优解

Author: sl1673495

算法/被围绕的区域/index.md

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+# 社区看到的优解
+
+其实这题本身是我想复杂了，我是一个个格子去遍历，然后再上下左右去扩展延伸。
+
+但是其实只需要遍历四个边界上的节点，遇到 O 的边界点才开始蔓延遍历，并且把遍历到的节点都标记为 M（防止重复遍历）
+
+最后再一次性遍历整个二维数组，遇到 M 的标记都转为 O（因为是从边界蔓延的，一定是不符合 X 的条件的）。
+
+这样遍历所走的路就会少很多。
+
+```js
+var solve = function (board) {
+  if (board.length == 0) return null;
+
+  for (var y = 0; y < board.length; y++) {
+    for (var x = 0; x < board[0].length; x++) {
+      if (board[y][x] == "O" && (y == 0 || y == board.length - 1 || x == 0 || x == board[0].length - 1)) {
+        dfs(board, y, x);
+      }
+    }
+  }
+
+  for (var y = 0; y < board.length; y++) {
+    for (var x = 0; x < board[0].length; x++) {
+      if (board[y][x] == "W") {
+        board[y][x] = "O";
+      } else {
+        board[y][x] = "X";
+      }
+    }
+  }
+
+  return board;
+};
+
+function dfs(board, y, x) {
+  if (y < 0 || x < 0 || y >= board.length || x >= board[0].length || board[y][x] == "X" || board[y][x] == "W") {
+    return;
+  }
+  board[y][x] = "W";
+  dfs(board, y + 1, x);
+  dfs(board, y - 1, x);
+  dfs(board, y, x + 1);
+  dfs(board, y, x - 1);
+  return;
+}
+```

算法/被围绕的区域-循环版.js renamed to 算法/被围绕的区域/被围绕的区域-循环版.js

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+var solve = function (board) {
+  if (board.length == 0) return null;
+
+  for (var y = 0; y < board.length; y++) {
+    for (var x = 0; x < board[0].length; x++) {
+      if (board[y][x] == "O" && (y == 0 || y == board.length - 1 || x == 0 || x == board[0].length - 1)) {
+        dfs(board, y, x);
+      }
+    }
+  }
+
+  for (var y = 0; y < board.length; y++) {
+    for (var x = 0; x < board[0].length; x++) {
+      if (board[y][x] == "W") {
+        board[y][x] = "O";
+      } else {
+        board[y][x] = "X";
+      }
+    }
+  }
+
+  return board;
+};
+
+function dfs(board, y, x) {
+  if (y < 0 || x < 0 || y >= board.length || x >= board[0].length || board[y][x] == "X" || board[y][x] == "W") {
+    return;
+  }
+  board[y][x] = "W";
+  dfs(board, y + 1, x);
+  dfs(board, y - 1, x);
+  dfs(board, y, x + 1);
+  dfs(board, y, x - 1);
+  return;
+}
+
+console.log(
+  solve([
+    ["X", "X", "X", "X"],
+    ["X", "O", "O", "X"],
+    ["X", "X", "O", "X"],
+    ["X", "O", "X", "X"],
+  ])
+);