トピック-島の数
これを中国のウェブサイトのタイトル（leetcode-cn）のアドレスに追加するだけです
与えられた2次元で構成されるグリッド'1'（土地）および'0'（水）、グリッド内の島の数をカウントしてください。島は常に水に囲まれており、各島は隣接する土地を水平方向または垂直方向に接続することによってのみ形成できます。
さらに、グリッドの4つの側面すべてが水に囲まれていると想定できます。
例1：
输入:
11110
11010
11000
00000
输出: 1
例2：
输入:
11000
11000
00100
00011
输出: 3
解释: 每座岛屿只能由水平和/或竖直方向上相邻的陆地连接而成。
方法：深さ優先探索（ShenDaoは1-> 0と考えた）、幅優先探索、複合検索
深さ優先探索（中国の公式）
class Solution {
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int nr = grid.length;
int nc = grid[0].length;
int num_islands = 0;
for (int r = 0; r < nr; ++r) {
for (int c = 0; c < nc; ++c) {
if (grid[r][c] == '1') {
++num_islands;
dfs(grid, r, c);
}
}
}
return num_islands;
}
}
void dfs(char[][] grid, int r, int c) {
int nr = grid.length;
int nc = grid[0].length;
if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
return;
}
grid[r][c] = '0';
dfs(grid, r - 1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
英語の公式ウェブサイトのほとんどの投票
- Python
def numIslands(self, grid):
if not grid:
return 0
count = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == '1':
self.dfs(grid, i, j)
count += 1
return count
def dfs(self, grid, i, j):
if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] != '1':
return
grid[i][j] = '#'
self.dfs(grid, i+1, j)
self.dfs(grid, i-1, j)
self.dfs(grid, i, j+1)
self.dfs(grid, i, j-1)
- Java
public class Solution {
private int n;
private int m;
public int numIslands(char[][] grid) {
int count = 0;
n = grid.length;
if (n == 0) return 0;
m = grid[0].length;
for (int i = 0; i < n; i++){
for (int j = 0; j < m; j++)
if (grid[i][j] == '1') {
DFSMarking(grid, i, j);
++count;
}
}
return count;
}
private void DFSMarking(char[][] grid, int i, int j) {
if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != '1') return;
grid[i][j] = '0';
DFSMarking(grid, i + 1, j);
DFSMarking(grid, i - 1, j);
DFSMarking(grid, i, j + 1);
DFSMarking(grid, i, j - 1);
}
- 注釈付き
public class Solution {
int y; // The height of the given grid
int x; // The width of the given grid
char[][] g; // The given grid, stored to reduce recursion memory usage
/**
* Given a 2d grid map of '1's (land) and '0's (water),
* count the number of islands.
*
* This method approaches the problem as one of depth-first connected
* components search
* @param grid, the given grid.
* @return the number of islands.
*/
public int numIslands(char[][] grid) {
// Store the given grid
// This prevents having to make copies during recursion
g = grid;
// Our count to return
int c = 0;
// Dimensions of the given graph
y = g.length;
if (y == 0) return 0;
x = g[0].length;
// Iterate over the entire given grid
for (int i = 0; i < y; i++) {
for (int j = 0; j < x; j++) {
if (g[i][j] == '1') {
dfs(i, j);
c++;
}
}
}
return c;
}
/**
* Marks the given site as visited, then checks adjacent sites.
*
* Or, Marks the given site as water, if land, then checks adjacent sites.
*
* Or, Given one coordinate (i,j) of an island, obliterates the island
* from the given grid, so that it is not counted again.
*
* @param i, the row index of the given grid
* @param j, the column index of the given grid
*/
private void dfs(int i, int j) {
// Check for invalid indices and for sites that aren't land
if (i < 0 || i >= y || j < 0 || j >= x || g[i][j] != '1') return;
// Mark the site as visited
g[i][j] = '0';
// Check all adjacent sites
dfs(i + 1, j);
dfs(i - 1, j);
dfs(i, j + 1);
dfs(i, j - 1);
}
}