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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());
}
};
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