題目連接
https://leetcode-cn.com/problems/number-of-islands/
題目描述
給定一個(gè)由 '1'(陸地)和 '0'(水)組成的的二維網(wǎng)格重慢,計(jì)算島嶼的數(shù)量。一個(gè)島被水包圍肄梨,并且它是通過(guò)水平方向或垂直方向上相鄰的陸地連接而成的欧啤。你可以假設(shè)網(wǎng)格的四個(gè)邊均被水包圍断凶。
示例 1:
輸入:
11110
11010
11000
00000
輸出: 1
示例 2:
輸入:
11000
11000
00100
00011
輸出: 3
go語(yǔ)言題解
深度優(yōu)先搜索
func numIslands(grid [][]byte) int {
rowsize := len(grid)
if rowsize <= 0{
return 0
}
colsize := len(grid[0])
num := 0
for i:=0; i<rowsize; i++{
for j:=0; j<colsize; j++{
if grid[i][j] == 49{
num++
dfs(grid, i, j)
}
}
}
return num
}
func dfs(grid[][] byte, r int, c int) {
rowsize := len(grid)
colsize := len(grid[0])
//標(biāo)記visited
grid[r][c] = 48
//如果有下一個(gè)節(jié)點(diǎn),并且節(jié)點(diǎn)沒(méi)有被訪問(wèn)過(guò),就深搜下一個(gè)節(jié)點(diǎn)
if r - 1 >= 0 && grid[r-1][c] == 49{
dfs(grid, r-1, c)
}
if r + 1 < rowsize && grid[r+1][c] == 49{
dfs(grid, r+1, c)
}
if c - 1 >= 0 && grid[r][c-1] == 49{
dfs(grid, r, c-1)
}
if c + 1 < colsize && grid[r][c+1] == 49{
dfs(grid, r, c+1)
}
}
廣度優(yōu)先搜索
func bfs(grid[][] byte) int{
rowsize := len(grid)
if rowsize <= 0{
return 0
}
colsize := len(grid[0])
num := 0
queue := make([]rc,0)
for r:=0; r<rowsize; r++{
for c:=0; c<colsize; c++{
//如果為 1 , 又沒(méi)有被訪問(wèn), 加入隊(duì)列
if grid[r][c] == 49{
//標(biāo)記visited
grid[r][c] = 48
queue = append(queue, rc{r, c})
num ++
//隊(duì)列沒(méi)有空說(shuō)明還有島嶼連接缅茉,可以繼續(xù)搜下一層
//如果隊(duì)列為空了凤巨,說(shuō)明這個(gè)島嶼也到了邊界了视乐,退出循環(huán),開(kāi)始搜下一個(gè)島嶼
for len(queue) != 0{
q := queue[0]
queue = queue[1:] //pop )
//搜索下一層敢茁,如果有沒(méi)有被訪問(wèn)過(guò)并值為1的炊林,則繼續(xù)加入隊(duì)列中
if q.r+1 < rowsize && grid[q.r+1][q.c] == 49{
grid[q.r+1][q.c] = 48
queue = append(queue, rc{q.r+1, q.c})
}
if q.r-1 >= 0 && grid[q.r-1][q.c] == 49{
grid[q.r-1][q.c] = 48
queue = append(queue, rc{q.r-1, q.c})
}
if q.c+1 < colsize && grid[q.r][q.c+1] == 49{
grid[q.r][q.c+1] = 48
queue = append(queue, rc{q.r, q.c+1})
}
if q.c-1 >=0 && grid[q.r][q.c-1] == 49{
grid[q.r][q.c-1] = 48
queue = append(queue, rc{q.r, q.c-1})
}
}
}
}
}
return num
}