Leetcode 200. Number of Islands
Given a 2d grid map of '1’s (land) and '0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input:
11110
11010
11000
00000
Output: 1
Example 2:
Input:
11000
11000
00100
00011
Output: 3
solution
深度优先搜索,遇到1的时候计数,并把其周围的数字1改为0
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
if (grid.empty())
return 0;
int n = grid.size();
int m = grid[0].size();
int ans = 0;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
{
if (grid[i][j] == '1')
{
ans++;
DFS(grid, i, j, n, m);
}
}
return ans;
}
void DFS(vector<vector<char>>& grid, int i, int j, int n, int m)
{
if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] == '0')
return;
grid[i][j] = '0';
DFS(grid, i + 1, j, n, m);
DFS(grid, i - 1, j, n, m);
DFS(grid, i, j + 1, n, m);
DFS(grid, i, j - 1, n, m);
}
};
本文详细解析了LeetCode200题“Number of Islands”的深度优先搜索算法解决方案,介绍了如何通过遍历二维网格,计算由'1'(陆地)构成的岛屿数量。通过递归调用深度优先搜索,将相连的陆地标记为已访问,避免重复计数。
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