200. Number of Islands

最新推荐文章于 2024-01-21 17:08:30 发布

卡其1991

最新推荐文章于 2024-01-21 17:08:30 发布

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本文介绍了一种使用递归深度优先搜索算法解决二维网格中岛屿数量计算问题的方法。通过遍历每个单元格并标记已访问过的陆地，确保了每个岛屿只被计数一次。

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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Answer: 1

Example 2:

Answer: 3

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

使用递归的深度优先算法

class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
int ret=0;
if(grid.empty()||grid[0].empty())
return ret;
int m=grid.size();
int n=grid[0].size();
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(grid[i][j]=='1')
{
dfs(grid,i,j,m,n);
ret++;
}
}
}
return ret;
}
void dfs(vector<vector<char>> &grid,int i,int j,int m,int n)
{
grid[i][j]='0';
if(i>0&&grid[i-1][j]=='1')
dfs(grid,i-1,j,m,n);
if(i<m-1&&grid[i+1][j]=='1')
dfs(grid,i+1,j,m,n);
if(j>0&&grid[i][j-1]=='1')
dfs(grid,i,j-1,m,n);
if(j<n-1&&grid[i][j+1]=='1')
dfs(grid,i,j+1,m,n);
}
};