11.14算法

发布时间 2023-11-14 11:26:56作者: 卧龙丹心

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> &grid)
{
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> &grid)
{
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