200. Number of Islands**(岛屿数量)

200. Number of Islands**(岛屿数量)

https://leetcode.com/problems/number-of-islands/

题目描述

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is '0' or '1'.

C++ 实现 1

可以作为学习 DFS 的练习题. 按 DFS 的方式遍历 grid, 如果遇到 '1', 那么进行岛屿数量的统计, 同时在 DFS 的过程中将遇到的 '1' 均置为 '0', 防止重复计算.

class Solution {
    
    
public:
    int numIslands(vector<vector<char>>& grid) {
    
    
        if (grid.empty() || grid[0].empty())
            return 0;

        int res = 0;
        for (int i = 0; i < grid.size(); ++i) {
    
    
            for (int j = 0; j < grid[0].size(); ++j) {
    
    
                if (grid[i][j] == '1') {
    
    
                    res ++;
                    dfs(grid, i, j);
                }
            }
        }
        return res;
    }

    void dfs(vector<vector<char>> &grid, int i, int j) {
    
    
        if (i < 0 || i >= grid.size() ||
            j < 0 || j >= grid[0].size() ||
            grid[i][j] == '0')
            return;

        grid[i][j] = '0';
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
};