[leetcode] Number of Islands-优快云博客

本文详细阐述了如何通过编程算法来计算二维网格中岛屿的数量，包括两种不同的解决方案：深度优先搜索（DFS）和广度优先搜索（BFS）。通过实例演示了如何遍历网格并标记已访问的陆地格子，最终计算出岛屿总数。

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From : https://leetcode.com/problems/number-of-islands/

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

Solution 1:

class Solution {
public:
    vector<vector<bool>> initCheck(vector<vector<char>>& grid) {
        vector<vector<bool>> check;
        int gsize = grid.size();
        for(int i=0; i<gsize; i++) {
            vector<bool> cur(grid[i].size(), false);
            check.push_back(cur);
        }
        return check;
    }
    
    void update(vector<vector<char>>& grid, vector<vector<bool>>& check, int i, int j) {
        check[i][j] = true;
        if(i-1>=0 && j<grid[i-1].size() && !check[i-1][j] && grid[i-1][j]=='1') update(grid, check, i-1, j);
        if(i+1<grid.size() && j<grid[i+1].size() && !check[i+1][j] && grid[i+1][j]=='1') update(grid, check, i+1, j);
        if(j-1>=0 && !check[i][j-1] && grid[i][j-1]=='1') update(grid, check, i, j-1);
        if(j+1<grid[i].size() && !check[i][j+1] && grid[i][j+1]=='1') update(grid, check, i, j+1);
    }
    
    int numIslands(vector<vector<char>>& grid) {
        vector<vector<bool>> check = initCheck(grid);
        int gsize = grid.size(), lsize, num=0;
        for(int i=0; i<gsize; i++) {
            lsize = grid[i].size();
            for(int j=0; j<lsize; j++) {
                if(!check[i][j] && grid[i][j]=='1') {
                    update(grid, check, i, j);
                    num++;
                }
            }
        }
        return num;
    }
};

Solution 2:

class Solution {
public:
    void dfs(vector<vector<char>> &grid, int x, int y) {
        if (x < 0 || x >= grid.size()) return;
        if (y < 0 || y >= grid[0].size()) return;
        if (grid[x][y] != '1') return;
        grid[x][y] = 'X';
        dfs(grid, x + 1, y);
        dfs(grid, x - 1, y);
        dfs(grid, x, y + 1);
        dfs(grid, x, y - 1);
    }
    
    int numIslands(vector<vector<char>> &grid) {
        if (grid.empty() || grid[0].empty()) return 0;
        int N = grid.size(), M = grid[0].size();
        int cnt = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j);
                    ++cnt;
                }
            }
        }
        return cnt;
    }
};