Leetcode Number of Islands

最新推荐文章于 2019-12-16 15:50:17 发布

martin_liang

最新推荐文章于 2019-12-16 15:50:17 发布

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本文介绍了一种使用深度优先搜索(DFS)算法来计算二维网格中岛屿数量的方法。该算法通过遍历每个陆地单元并标记已访问过的陆地来实现。文章提供了完整的C++代码实现，并通过两个示例说明了如何正确计算出岛屿的数量。

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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

class Solution {
public:

    void DFSVisit(vector<vector<char>> &grid, int i, int j, vector<vector<bool>> &visited)
    {
        int rows = grid.size();
        int coloms = grid[0].size();
        
        int horizontal[4] = {-1, 0, 1, 0};
        int vertical[4]   = {0, -1, 0, 1};
        
        visited[i][j] = true;
        
        for (int idx=0; idx<4; idx++)
        {
            int new_i = i+horizontal[idx];
            int new_j = j+vertical[idx];
            if (new_i>=0 && new_i<rows && new_j>=0 && new_j<coloms)
            {
                if (grid[new_i][new_j]=='1' && !visited[new_i][new_j])
                    DFSVisit(grid, new_i, new_j, visited);
            }
        }
    }

    int numIslands(vector<vector<char>> &grid) {
        
        if (grid.size()==0 || grid[0].size() ==0)
            return 0;
        
        int rows = grid.size();
        int colums = grid[0].size();
        
        vector<vector<bool>> visited(rows, vector<bool>(colums));
        
        for (int i=0; i<rows; i++)
            for (int j=0; j<colums; j++)
                visited[i][j] = false;
                
        int count = 0;
        for (int i=0; i<rows; i++)
        {
            for (int j=0; j<colums; j++)
            {
                if (grid[i][j]=='1' && !visited[i][j])
                {
                    DFSVisit(grid, i, j, visited);
                    count++;
                }
            }
        }
        
        return count;
    }
};