Leetcode 200.岛屿的数量 - DFS、BFS

标签：函数调用   return   坐标   air   pop   入队   pre   class   make   

Leetcode 200 岛屿的数量：

DFS利用函数调用栈保证了检索顺序，

BFS则需要自己建立队列，把待检索对象按规则入队。

class Solution {  // DFS解法，8ms/10.7MB，99.7% / 92%
public:

    /**
    * row,col: 坐标，以0开始.
    * 用row和col，而不是x，y. 否则容易写成grid[x][y],顺序不对！！
    */
    void dfs_set_zero(vector<vector<char>>& grid, int row, int col) {
        int n_rows = grid.size();
        if (!n_rows) return;
        int n_cols = grid[0].size();
        if (!n_cols) return;
        
        if (row<0 || row>=n_rows || col<0 || col>=n_cols) return;
        
        if (grid[row][col] == '1') {
            grid[row][col] = '0';
            dfs_set_zero(grid, row-1, col);
            dfs_set_zero(grid, row+1, col);
            dfs_set_zero(grid, row, col-1);
            dfs_set_zero(grid, row, col+1);
        }
    }

    int numIslands(vector<vector<char>>& grid) {
        int result = 0;
        for (int i=0; i<grid.size(); ++i) {
            for (int j=0; j<grid[0].size(); ++j) {
                if (grid[i][j] == '1') {
                    result++;
                    dfs_set_zero(grid, i, j);
                }
            }
        }
        return result;
    }
};

class Solution {  // BFS解法，12ms/11MB, 96.9% / 69.4%
public:
    queue<pair<int,int>> q;
    
    void bfs_set_zero(vector<vector<char>>& grid) {
        int n_rows = grid.size();
        if(!n_rows) return;
        int n_cols = grid[0].size();
        if(!n_cols) return;
        
        while (!q.empty()) {
            auto pos = q.front();
            q.pop();
            if (grid[pos.first][pos.second] == '1') {
                grid[pos.first][pos.second] = '0';
                if (pos.first-1>=0 && grid[pos.first-1][pos.second] == '1')
                    q.emplace(pos.first-1,pos.second);
                if (pos.first+1<n_rows && grid[pos.first+1][pos.second] == '1')
                    q.emplace(pos.first+1,pos.second);
                if (pos.second-1>=0 && grid[pos.first][pos.second-1] == '1')
                    q.emplace(pos.first,pos.second-1);
                if (pos.second+1<n_cols && grid[pos.first][pos.second+1] == '1')
                    q.emplace(pos.first,pos.second+1);
            }
        }
    }
    
    int numIslands(vector<vector<char>>& grid) {
        int result = 0;
        for (int i=0; i<grid.size(); i++) {
            for (int j=0; j<grid[0].size(); j++) {
                if (grid[i][j] == '1') {
                    result++;
                    //q.push(make_pair(i,j));
                    q.emplace(i,j);  // 使用emplace而非push，constructed in-place
                    bfs_set_zero(grid);
                }
            }
        }
        return result;
    }
};

Leetcode 200.岛屿的数量 - DFS、BFS

标签：函数调用   return   坐标   air   pop   入队   pre   class   make   

原文地址：https://www.cnblogs.com/dylanchu/p/11326423.html