30 Day Challenge Day 8 | Leetcode 200. Number of Islands

题解

这道题做了若干遍，推荐的做法是drown every island的做法，也就是用初始的grid矩阵同时作为记忆矩阵，那么可以省去一个visited矩阵。这里作为DFS练习，还是套用visited矩阵的做法。

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        if(grid.empty()) return 0;
        int num = 0;
        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), false));
        for(int i = 0; i < grid.size(); i++) {
            for(int j = 0; j < grid[0].size(); j++) {
                if(grid[i][j] == '1' && !visited[i][j]) {
                    search(grid, visited, i, j);
                    num++;
                }
            }
        }
        return num;
    }
    
    void search(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
        if(x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size()) return;
        if(grid[x][y] == '0') return;
        if(visited[x][y]) return;

        visited[x][y] = true;
        
        int dirs[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        
        for(int k = 0; k < 4; k++) {
            int xx = x + dirs[k][0];
            int yy = y + dirs[k][1];
            search(grid, visited, xx, yy);
        }
    }
};