图论

其实图是什么我都不知道…

200.岛屿数量

深度优先算法和广度优先算法

应该蛮好理解的吧…

DFS解法

深度优先搜索

class Solution {
public:
    int m,n;
    void dfs(vector<vector<char>>& grid,int x,int y){
        if(x<0||x>=m||y<0||y>=n||grid[x][y]=='0') return; //越界 or 遇到水 直接停
        grid[x][y] = '0'; //把当前陆地淹掉
        dfs(grid,x+1,y);
        dfs(grid,x-1,y);  //四个方向递归
        dfs(grid,x,y+1);
        dfs(grid,x,y-1);
    }
    int numIslands(vector<vector<char>>& grid) {
        m = grid.size();
        n = grid[0].size();
        int count = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == '1'){
                    count++;  //发现一个新岛
                    dfs(grid,i,j); //淹掉整个岛屿
                }
            }
        }
        return count;
    }
};

BFS解法

广度优先搜索

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int count = 0;
        int dx[4] = {0, 0, 1, -1};
        int dy[4] = {1, -1, 0, 0};
        queue<pair<int, int>> q;  //队列
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    grid[i][j] = '0';
                    q.push({i, j});
                    while (!q.empty()) {
                        auto [x, y] = q.front();
                        q.pop();
                        for (int k = 0; k < 4; k++) {
                            int nx = x + dx[k];
                            int ny = y + dy[k];
                            if (nx >= 0 && nx < m && ny >= 0 && ny < n &&
                                grid[nx][ny] == '1') {
                                grid[nx][ny] = '0';
                                q.push({nx, ny});
                            }
                        }
                    }
                }
            }
        }
        return count;
    }
};