Leetcode Hot 100 200.岛屿数量
1.题目
200. 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
2.代码及解析
和之前那个岛屿面积差不多其实 但是我还是看答案了 还是不会
class Solution {
int count=0;
void getArea(vector<vector<char>>& grid, int i, int j)
{
//由于坐标每次 +1 ,所以判断是否等于数组长度即可
if (i == grid.size() || i < 0) //横坐标是否超过最大值 或者为负数
return ;
else if (j == grid[0].size() || j < 0) //纵坐标是否超过最大值 或者为负数
return ;
if (grid[i][j] == '1') //是否为陆地,是的话执行If
{
grid[i][j] = '0'; //将其置为0
getArea(grid, i + 1, j) ;
getArea(grid, i - 1, j ) ;
getArea(grid, i, j + 1) ;
getArea(grid, i, j - 1); //递归 把四边周围都加起来
} //如果是0,说明是海,不是岛屿 不需要再看周围了。
}
public:
int numIslands(vector<vector<char>>& grid) {
for (int i = 0; i < grid.size(); i++) //每一个点都遍历一遍求岛屿面积
{
for (int j = 0; j < grid[0].size(); j++)
{
if (grid[i][j] == '1')
{
count++;
//以此为中心,向四周扩
getArea(grid, i, j);
}
}
}
return count;
}
};