[路飞]_前端算法第八十五弹-岛屿数量

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

这道题我们使用深度优先遍历，我们从[[0][0]]开始遍历，遇到陆地”1”，我们就遍历其周边地区，如果都是1，我们将其赋值为-1，代表遍历过，对总数+1，再继续遍历没有遍历过的地图区域。直到全部遍历完成。

记得遍历过程要注意边界条件，遍历不能超过地图范围。

var numIslands = function (grid) {
  let len = grid.length;
  let row = grid[0].length
  let res = 0;
  for (let i = 0; i < len; i++) {
    for (let j = 0; j < row; j++) {
      if (grid[i][j] == 1) {
        dfs(i, j)
        res++
      }
    }
  }
  function dfs (i, j) {
    if (grid[i][j] == 1) {
      grid[i][j] = -1
      if (i + 1 < len) dfs(i + 1, j)
      if (j + 1 < row) dfs(i, j + 1)
      if (i > 0) dfs(i - 1, j)
      if (j > 0) dfs(i, j - 1)
    }
  }
  return res
};