【一天一大 lee】岛屿的周长 (难度:简单) - Day20201030

输入:
[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

输出: 16

解释: 它的周长是下面图片中的 16 个黄色的边：

/**
 * @param {number[][]} grid
 * @return {number}
 */
var islandPerimeter = function(grid) {
  let m = grid.length,
    n = grid[0] ? grid[0].length : 0,
    _result = 0
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === 1) {
        if (i == 0 || grid[i - 1][j] == 0) _result++;
        if (i == m - 1 || grid[i + 1][j] == 0) _result++;
        if (j == 0 || grid[i][j - 1] == 0) _result++;
        if (j == n - 1 || grid[i][j + 1] == 0) _result++;
      }
    }
  }
  return _result
};

/**
 * @param {number[][]} grid
 * @return {number}
 */
var islandPerimeter = function(grid) {
  let m = grid.length,
    n = grid[0] ? grid[0].length : 0,
    _result = 0
  if (n === 0 || m === 0) return _result

  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === 1) {
        _result += dfs(i, j);
      }
    }
  }

  function dfs (x, y) {
    // 与边界相邻
    if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] === 0) return 1
    // 非边界的1检查是否计算过
    if (grid[x][y] === 2) return 0
    grid[x][y] = 2;
    return dfs(x - 1, y) + dfs(x + 1, y) + dfs(x, y - 1) + dfs(x, y + 1)
  }
  return _result
};