200\. 岛屿数量
# 200. 岛屿数量 (opens new window)
# Description
Difficulty: 中等
Related Topics: 深度优先搜索 (opens new window), 广度优先搜索 (opens new window), 并查集 (opens new window), 数组 (opens new window), 矩阵 (opens new window)
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
1
2
3
4
5
6
7
2
3
4
5
6
7
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
1
2
3
4
5
6
7
2
3
4
5
6
7
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j]
的值为'0'
或'1'
# Solution
Language: JavaScript
var numIslands = function(grid) {
function dfs(x, y) {
grid[x][y] = 0
if(grid[x - 1] && grid[x - 1][y] == 1) dfs(x-1, y)
if(grid[x + 1] && grid[x + 1][y] == 1) dfs(x+1, y)
if(grid[x][y - 1] == 1) dfs(x, y - 1)
if(grid[x][y + 1] == 1) dfs(x, y + 1)
}
let count = 0
for(let i = 0; i < grid.length; i++) {
for(let j = 0; j< grid[0].length; j++) {
if(grid[i][j] == 1) {
dfs(i, j)
count++
}
}
}
return count
};
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20
2
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20