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
1、dfs
#include <vector> #include <iostream> #include <queue> using namespace std; class Solution { public: int numIslands(vector<vector<char>> &grid) { int N = grid.size(); int M = grid[0].size(); vector<vector<bool>> visited(N, vector<bool>(M, false)); int res = 0; queue<pair<int, int>> qe; for (int i = 0; i < N; ++i) { for (int j = 0; j < M; ++j) { if (!visited[i][j] && grid[i][j] != ‘0‘) { qe.push({i, j}); res++; } while (!qe.empty()) { auto ele = qe.front(); qe.pop(); if ((ele.first + 1) < N && !visited[ele.first + 1][ele.second] && grid[ele.first + 1][ele.second] != ‘0‘) { qe.push({ele.first + 1, ele.second}); visited[ele.first + 1][ele.second] = true; } if ((ele.second + 1) < M && !visited[ele.first][ele.second + 1] && grid[ele.first][ele.second + 1] != ‘0‘) { qe.push({ele.first, ele.second + 1}); visited[ele.first][ele.second + 1] = true; } if ((ele.first - 1) > -1 && !visited[ele.first - 1][ele.second] && grid[ele.first - 1][ele.second] != ‘0‘) { qe.push({ele.first - 1, ele.second}); visited[ele.first - 1][ele.second] = true; } if ((ele.second - 1) > -1 && !visited[ele.first][ele.second - 1] && grid[ele.first][ele.second - 1] != ‘0‘) { qe.push({ele.first, ele.second - 1}); visited[ele.first][ele.second - 1] = true; } } } } return res; } }; int main() { vector<vector<char>> grid{{‘1‘, ‘1‘, ‘1‘}, {‘0‘, ‘1‘, ‘0‘}, {‘1‘, ‘1‘, ‘1‘}}; Solution s; cout << s.numIslands(grid) << endl; }
原文:https://www.cnblogs.com/zhangzhangtabszj/p/14443728.html