题目
岛屿数量
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'
答案
class Solution {
public:
void dfs(int i, int j, vector<vector
{
if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size() || grid[i][j] == '0')
{
return;
}
grid[i][j] = '0';
dfs(i - 1, j, grid);
dfs(i, j - 1, grid);
dfs(i, j + 1, grid);
dfs(i + 1, j, grid);
}
int numIslands(vector<vector
{
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(i , j, grid);
}
}
}
return res;
}
};
关键
dfs，找到1个1，剩下的都让它为0就行
标签：int,res,11.14,dfs,++,算法,grid,size From： https://www.cnblogs.com/minipython-wldx/p/17831212.html