題目地址：Number of Islands
題目簡介：
給定一個由 '1'（陸地）和 '0'（水）組成的的二維網格，計算島嶼的數量。一個島被水包圍，並且它是通過水平方向或垂直方向上相鄰的陸地連接而成的。你可以假設網格的四個邊均被水包圍。
Example 1:
Input:
11110
11010
11000
00000
Output: 1
Example 2:
Input:
11000
11000
00100
00011
Output: 3
題目解析：
1、BFS
每第一次碰到'1'，向四周發散把所有的相鄰的'1'找到，一個都不留。
C++（利用queue）
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if(grid.size() == 0)
return 0;
int high = grid.size(), width = grid[0].size();
int ans = 0;
for (int i = 0; i < high; i++)
{
for (int j = 0; j < width; j++)
{
if (grid[i][j] == '1')
{
ans++;
queue<int> pos;
pos.push(i), pos.push(j);
while(!pos.empty())
{
int x = pos.front();
pos.pop();
int y = pos.front();
pos.pop();
grid[x][y] = '0';
if (x + 1 < high && grid[x + 1][y] == '1')
{
pos.push(x + 1), pos.push(y);
grid[x + 1][y] = '0';
}
if (y + 1 < width && grid[x][y + 1] == '1')
{
pos.push(x), pos.push(y + 1);
grid[x][y + 1] = '0';
}
if (x - 1 > -1 && grid[x - 1][y] == '1')
{
pos.push(x - 1), pos.push(y);
grid[x - 1][y] = '0';
}
if (y - 1 > -1 && grid[x][y - 1] == '1')
{
pos.push(x), pos.push(y - 1);
grid[x][y - 1] = '0';
}
}
}
}
}
return ans;
}
};
Python：
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if len(grid) == 0:
return 0
high = len(grid)
width = len(grid[0])
ans = 0
for i in range(high):
for j in range(width):
if grid[i][j] == '1':
ans += 1
pos = []
pos.append(j)
pos.append(i)
while(len(pos)):
x = pos.pop()
y = pos.pop()
grid[x][y] = '0'
if (x + 1 < high and grid[x + 1][y] == '1'):
pos.append(y)
pos.append(x + 1)
grid[x + 1][y] = '0'
if (y + 1 < width and grid[x][y + 1] == '1'):
pos.append(y + 1)
pos.append(x)
grid[x][y + 1] = '0'
if (x - 1 > -1 and grid[x - 1][y] == '1'):
pos.append(y)
pos.append(x - 1)
grid[x - 1][y] = '0'
if (y - 1 > -1 and grid[x][y - 1] == '1'):
pos.append(y - 1)
pos.append(x)
grid[x][y - 1] = '0'
return ans
2、DFS
每次碰到第一次的'1'，就先朝一個方向把'1'找完，一個都不剩。
C++：
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if (grid.empty())
return 0;
int high = grid.size(), width = grid[0].size();
int ans = 0;
for (int i = 0; i < high; i++)
{
for (int j = 0; j < width; j++)
{
if (grid[i][j] == '0')
continue;
helper(grid, i, j);
ans++;
}
}
return ans;
}
void helper(vector<vector<char>>& grid, int x, int y) {
if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == '0')
return;
grid[x][y] = '0';
helper(grid, x - 1, y);
helper(grid, x + 1, y);
helper(grid, x, y - 1);
helper(grid, x, y + 1);
}
};
Python：
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if len(grid) == 0:
return 0
high = len(grid)
width = len(grid[0])
ans = 0
for i in range(high):
for j in range(width):
if grid[i][j] == '0':
continue
self.helper(grid, i, j)
ans += 1
return ans
def helper(self, grid, x, y):
if (x < 0 or x >= len(grid) or y < 0 or y >= len(grid[0]) or grid[x][y] == '0'):
return
grid[x][y] = '0'
self.helper(grid, x - 1, y)
self.helper(grid, x + 1, y)
self.helper(grid, x, y - 1)
self.helper(grid, x, y + 1)