BFS 遍历例子

以Leetcode题目 200. 岛屿数量 为例 展示BFS代码
题目如下 https://leetcode-cn.com/problems/number-of-islands/

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外，你可以假设该网格的四条边均被水包围。
 
示例 1：
输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1
 
示例 2：
输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

BFS遍历

class Solution {
public:
    int addX[4] = {1,0,0,-1};
    int addY[4] = {0,1,-1,0};
     
    void bfs(vector<vector<char>>& grid,int x,int y){
        grid[x][y] = '0';
        queue<pair<int,int>> q;
        q.push({x,y});
         
        while(!q.empty()){
            int x = q.front().first;
            int y = q.front().second;
            q.pop();
             
            for(int i = 0; i < 4;i++){
                int newx = x+addX[i];
                int newy = y+addY[i];
                 
                if(newx>=0 && newx < grid.size() && newy>=0 && newy<grid[0].size() && grid[newx][newy] == '1'){
                    grid[newx][newy] = '0';
                    q.push({newx,newy});
                }
            }
        }
         
    }
     
    int numIslands(vector<vector<char>>& grid) {
        int ans  =0;
        for(int i = 0; i < grid.size();i++){
            for(int j = 0; j < grid[0].size();j++){
                if(grid[i][j] == '1'){
                    bfs(grid,i,j); ans++;
                }     
            }
        }
         
        return ans;
    }
};

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