René's URL Explorer Experiment

{"@context":"https://schema.org","@type":"DiscussionForumPosting","headline":"1254.统计封闭岛屿的数目","articleBody":"[原题地址](https://leetcode-cn.com/problems/number-of-closed-islands/)\r\n\u003e 有一个二维矩阵 grid ，每个位置要么是陆地（记号为 0 ）要么是水域（记号为 1 ）。\r\n我们从一块陆地出发，每次可以往上下左右 4 个方向相邻区域走，能走到的所有陆地区域，我们将其称为一座「岛屿」。\r\n如果一座岛屿 完全 由水域包围，即陆地边缘上下左右所有相邻区域都是水域，那么我们将其称为 「封闭岛屿」。\r\n请返回封闭岛屿的数目。\r\n\r\n### 题解\r\ndfs\r\n1. 判断当前点是否是陆地， grid[i][j] === 0\r\n2. 如果是陆地，res++, 将其置为1，深度优先搜索，将当前点相邻的所有点标记为1\r\n\r\n注意的是，要求是 **封闭岛屿**， 需要预先处理边界上的陆地，不算入结果数量\r\n\r\n### 代码\r\n\r\n```js\r\n/**\r\n * @param {number[][]} grid\r\n * @return {number}\r\n */\r\nvar closedIsland = function(grid) {\r\n  for (let i = 0; i \u003c grid[0].length; i++) {\r\n    dfs(grid, 0, i)\r\n    dfs(grid, grid.length - 1, i)\r\n  }\r\n  for (let i = 0; i \u003c grid.length; i++) {\r\n    dfs(grid, i, 0)\r\n    dfs(grid, i, grid[0].length - 1)\r\n  }\r\n  let res = 0\r\n  console.log(grid)\r\n  for (let i = 0; i \u003c grid.length; i++) {\r\n    for (let j = 0; j \u003c grid[0].length; j++) {\r\n      if (grid[i][j] === 0) {\r\n        res++\r\n        dfs(grid, i, j)\r\n      }\r\n    }\r\n  }\r\n  return res;\r\n};\r\n\r\nfunction dfs(grid, x, y) {\r\n  if (!isValid(grid, x, y) || grid[x][y] !== 0) {\r\n    return\r\n  }\r\n\r\n  grid[x][y] = 1;\r\n  dfs(grid, x - 1, y)\r\n  dfs(grid, x + 1, y)\r\n  dfs(grid, x, y - 1)\r\n  dfs(grid, x, y + 1)\r\n}\r\n\r\nfunction isValid(grid, x, y) {\r\n  return x \u003e= 0 \u0026\u0026 x \u003c grid.length \u0026\u0026 y \u003e= 0 \u0026\u0026 y \u003c grid[0].length\r\n}\r\n\r\n```\r\n\r\n\r\n\r\n\r\n\r\n\r\n","author":{"url":"https://github.com/feikerwu","@type":"Person","name":"feikerwu"},"datePublished":"2020-03-05T08:24:25.000Z","interactionStatistic":{"@type":"InteractionCounter","interactionType":"https://schema.org/CommentAction","userInteractionCount":2},"url":"https://github.com/8/algorithm-camp/issues/8"}