Preface
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The Problem
Given a 2D grid composed of '1' (land) and '0' (water), count the number of islands in the grid.
An island is always surrounded by water, and each island is formed only by connecting adjacent lands horizontally and/or vertically.
Additionally, you may assume all four edges of the grid are surrounded by water.
Example 1
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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
Example 2
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输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3
Implementation
Using depth-first search (DFS)
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class Solution {
private:
void dfs(vector<vector<char>>& grid, int r, int c) {
int nr = grid.size();
int nc = grid[0].size();
grid[r][c] = '0';
if (r - 1 >= 0 && grid[r-1][c] == '1') dfs(grid, r - 1, c);
if (r + 1 < nr && grid[r+1][c] == '1') dfs(grid, r + 1, c);
if (c - 1 >= 0 && grid[r][c-1] == '1') dfs(grid, r, c - 1);
if (c + 1 < nc && grid[r][c+1] == '1') dfs(grid, r, c + 1);
}
public:
int numIslands(vector<vector<char>>& grid) {
int nr = grid.size();
if (!nr) return 0;
int nc = grid[0].size();
int num_islands = 0;
for (int r = 0; r < nr; ++r) {
for (int c = 0; c < nc; ++c) {
if (grid[r][c] == '1') {
++num_islands;
dfs(grid, r, c);
}
}
}
return num_islands;
}
};