LeetCode 200-Number of Islands

分析

只需要对每一个陆地区域做一次dfs，每次dfs中将已经遍历的陆地网格“1”变为水域网格“0”（防止再次遍历导致重复）。对每次dfs计数，总共dfs的次数即为岛屿总数。应注意，grid为空的特殊情况应该排除。

class Solution {
public:
    int numIslands(vector> &grid) {
        int row = grid.size();
        if (!row) return 0;
        int col = grid[0].size();
        int count = 0;

        for (unsigned i = 0; i != row; ++i) {
            for (unsigned j = 0; j != col; ++j) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j); ++count;
                }
            }
        }

        return count;
    }
    void dfs(vector> &grid, unsigned i, unsigned j) {
        int dx[4] = {1, 0, -1, 0};
        int dy[4] = {0, 1, 0, -1};
        int row = grid.size(), col = grid[0].size();
        grid[i][j] = '0';

        for (int k = 0; k != 4; ++k) {
            int new_i = i + dx[k], new_j = j + dy[k];
            if (new_i >=0 && new_i < row && new_j >= 0 && new_j < col && grid[new_i][new_j] == '1')
                dfs(grid, new_i, new_j);
        }
    }
};