题目链接:https://leetcode-cn.com/problems/number-of-islands
题目描述:
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:grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出:1
示例 2:
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
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'
题解:
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
int ans = 0;
for(int i = 0; i < grid.size(); i++)
{
for(int j = 0; j < grid[0].size(); j++)
{
if(grid[i][j] == '1')
{
ans++;
dfs(grid, i, j);
}
}
}
return ans;
}
void dfs(vector<vector<char>>& grid, int r, int c)
{
//base 判断是否越界
if(!isLands(grid, r, c))
return;
if(grid[r][c] != '1')
return;
//标记已经访问过的点
grid[r][c] = '2';
//遍历周围点
dfs(grid, r - 1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
bool isLands(vector<vector<char>>& grid, int r, int c)
{
return (r >= 0 && r < grid.size() && c >= 0 && c <grid[0].size());
}
};