【Medium】200. Number of Islands

最新推荐文章于 2021-11-16 21:49:30 发布

原创最新推荐文章于 2021-11-16 21:49:30 发布 · 280 阅读

CC 4.0 BY-SA版权

本文介绍了一种使用深度优先搜索（DFS）结合递归的方法来计算二维网格中岛屿的数量。通过将‘1’标记为‘0’并检查相邻元素，确保所有岛屿被正确计数。

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

中心思想：

DFS+Recursion

如果一个格子是1，将其标为0后，查其上下左右，重复把查到的1标为0。当一个岛全为0后，recursion会停止，记住找到了一个岛，然后继续找地图中的下一个1.

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