1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
| // @Title: 岛屿数量 (Number of Islands)
// @Author: 15816537946@163.com
// @Date: 2019-11-19 15:04:41
// @Runtime: 0 ms
// @Memory: 2.8 MB
/*
* @lc app=leetcode.cn id=200 lang=golang
*
* [200] 岛屿数量
*
* https://leetcode-cn.com/problems/number-of-islands/description/
*
* algorithms
* Medium (45.46%)
* Likes: 222
* Dislikes: 0
* Total Accepted: 26.5K
* Total Submissions: 58.4K
* Testcase Example: '[["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]'
*
* 给定一个由 '1'(陆地)和
* '0'(水)组成的的二维网格,计算岛屿的数量。一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
*
* 示例 1:
*
* 输入:
* 11110
* 11010
* 11000
* 00000
*
* 输出: 1
*
*
* 示例 2:
*
* 输入:
* 11000
* 11000
* 00100
* 00011
*
* 输出: 3
*
*
*/
/*
func numIslands(grid [][]byte) int {
n := len(grid)
if n == 0 {
return 0
}
m := len(grid[0])
var dfs func(int, int)
dfs = func(i, j int) {
grid[i][j] = '0'
if 0 <= i-1 && grid[i-1][j] == '1' {
dfs(i-1, j)
}
if 0 <= j-1 && grid[i][j-1] == '1' {
dfs(i, j-1)
}
if i+1 < n && grid[i+1][j] == '1' {
dfs(i+1, j)
}
if j+1 < m && grid[i][j+1] == '1' {
dfs(i, j+1)
}
}
var ret int
for i := range grid {
for j := range grid[0] {
if grid[i][j] == '1' {
ret++
dfs(i, j)
}
}
}
return ret
}
*/
func numIslands(grid [][]byte) int {
n := len(grid)
if n == 0 {
return 0
}
m := len(grid[0])
var dfs func(int, int)
dfs = func(i, j int) {
grid[i][j] = '0'
if i > 0 && grid[i-1][j] == '1' {
dfs(i-1, j)
}
if j > 0 && grid[i][j-1] == '1' {
dfs(i, j-1)
}
if i < n-1 && grid[i+1][j] == '1' {
dfs(i+1, j)
}
if j < m-1 && grid[i][j+1] == '1' {
dfs(i, j+1)
}
}
var cnt int
for i :=range grid {
for j := range grid[0] {
if grid[i][j] == '1' {
cnt++
dfs(i, j)
}
}
}
return cnt
}
|