给定 '1'（陆地）和 '0'（水）的二维网格图，计算岛屿的数量。一个岛被水包围，并且通过水平或垂直连接相邻的陆地而形成。你可以假设网格的四个边均被水包围。
示例 1:
11110
11010
11000
00000
答案: 1
示例 2:
11000
11000
00100
00011
答案: 3
详见：https://leetcode.com/problems/number-of-islands/description/
Java实现：
class Solution {
public int numIslands(char[][] grid) {
int m=grid.length;
if(m==0||grid==null){
return 0;
}
int res=0;
int n=grid[0].length;
boolean[][] visited=new boolean[m][n];
for(int i=0;i<m;++i){
for(int j=0;j<n;++j){
if(grid[i][j]=='1'&&!visited[i][j]){
numIslandsDFS(grid,visited,i,j);
++res;
}
}
}
return res;
}
private void numIslandsDFS(char[][] grid,boolean[][] visited,int x,int y){
if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] != '1' || visited[x][y]){
return;
}
visited[x][y] = true;
numIslandsDFS(grid, visited, x - 1, y);
numIslandsDFS(grid, visited, x + 1, y);
numIslandsDFS(grid, visited, x, y - 1);
numIslandsDFS(grid, visited, x, y + 1);
}
}
C++实现：
class Solution {
public:
int numIslands(vector<vector<char> > &grid) {
if (grid.empty() || grid[0].empty())
{
return 0;
}
int m = grid.size(), n = grid[0].size(), res = 0;
vector<vector<bool> > visited(m, vector<bool>(n, false));
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
if (grid[i][j] == '1' && !visited[i][j])
{
numIslandsDFS(grid, visited, i, j);
++res;
}
}
}
return res;
}
void numIslandsDFS(vector<vector<char> > &grid, vector<vector<bool> > &visited, int x, int y) {
if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] != '1' || visited[x][y])
{
return;
}
visited[x][y] = true;
numIslandsDFS(grid, visited, x - 1, y);
numIslandsDFS(grid, visited, x + 1, y);
numIslandsDFS(grid, visited, x, y - 1);
numIslandsDFS(grid, visited, x, y + 1);
}
};