【LeetCode】200. 岛屿数量

题目

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:
11110
11010
11000
00000
输出: 1

示例 2:

输入:
11000
11000
00100
00011
输出: 3

解释: 每座岛屿只能由水平和/或竖直方向上相邻的陆地连接而成。

思路

代码

时间复杂度：O(row * col)
空间复杂度：O(row * col)

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        if (grid.empty()) return 0;
        int res = 0, row = grid.size(), col = grid[0].size();
        vector<vector<bool>> visited(row, vector<bool>(col));
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (grid[i][j] == '1' && visited[i][j] == false) {
                    dfs(grid, visited, i, j);
                    ++res;
                }                
            }
        }
        return res;
    }

    void dfs(vector<vector<char>> &grid, vector<vector<bool>> &visited, int i, int j) {
        int row = grid.size(), col = grid[0].size();
        if (i < 0 || i >= row || j < 0 || j >= col || grid[i][j] == '0' || visited[i][j] == true) return;
        visited[i][j] = true;
        dfs(grid, visited, i - 1, j);
        dfs(grid, visited, i + 1, j);
        dfs(grid, visited, i, j - 1);
        dfs(grid, visited, i, j + 1);
    }
};

优化空间复杂度

访问过就重置值为 0。

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {        
        if (grid.empty() || grid[0].empty()) return 0;
        int row = grid.size(), col = grid[0].size(), c = 0;
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (grid[i][j] == '1') {
                    ++c;
                    dfs(grid, i, j);
                }                
            }
        }
        return c;
    }
    
    void dfs(vector<vector<char>>& grid, int row, int col) {        
        if (row < 0 || row >= grid.size() || col < 0 || col >= grid[0].size() || grid[row][col] == '0') return;
        grid[row][col] = '0';
        dfs(grid, row - 1, col);
        dfs(grid, row + 1, col);
        dfs(grid, row, col - 1);
        dfs(grid, row, col + 1);
    }
};