链接:200. 岛屿数量
深度优先搜索
版本1
class Solution {
// 上下左右四个方向遍历
int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
// 记录行,列数
int m, n;
// 标记数组,判断是否已经访问过
boolean[][] visited = null;
public int numIslands(char[][] grid) {
if (grid == null || grid[0].length == 0) {
return 0;
}
int res = 0;
m = grid.length;
n = grid[0].length;
visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == '1' && !visited[i][j]) {
res++;
dfs(grid, i, j);
}
}
}
return res;
}
/**
* @param grid
* @param x
* @param y
* @return
*/
private boolean dfs(char[][] grid, int x, int y) {
// 如果当前遍历的字符是陆地,则继续访问后续位置
if (grid[x][y] == '1') {
visited[x][y] = true;
// 依次从四个方向分别计算
for (int i = 0; i < 4; i++) {
int newX = x + direction[i][0];
int newy = y + direction[i][1];
if (inArea(newX, newy) && !visited[newX][newy]) {
if (dfs(grid, newX, newy)) {
return true;
}
}
}
}
return false;
}
/**
* 判断当前坐标点是否越界
*
* @param x
* @param y
* @return
*/
private boolean inArea(int x, int y) {
return x >= 0 && x < m && y >= 0 && y < n;
}
}
版本2
class Solution {
// 上下左右四个方向遍历
int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
// 记录行,列数
int m, n;
public int numIslands(char[][] grid) {
if (grid == null || grid[0].length == 0) {
return 0;
}
int res = 0;
m = grid.length;
n = grid[0].length;
visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == '1') {
res++;
dfs(grid, i, j);
}
}
}
return res;
}
/**
* @param grid
* @param x
* @param y
* @return
*/
private boolean dfs(char[][] grid, int x, int y) {
// 如果当前遍历的字符是陆地,则继续访问后续位置
if (grid[x][y] == '1') {
// 一点小改进,直接将当前字符改为0,变相的设置为已访问,节省内存消耗,避免开辟visited内存
//visited[x][y] = true;
grid[x][y] = '0';
// 依次从四个方向分别计算
for (int i = 0; i < 4; i++) {
int newX = x + direction[i][0];
int newy = y + direction[i][1];
if (inArea(newX, newy) && grid[newX][newy] == '1') {
if (dfs(grid, newX, newy)) {
return true;
}
}
}
}
return false;
}
/**
* 判断当前坐标点是否越界
*
* @param x
* @param y
* @return
*/
private boolean inArea(int x, int y) {
return x >= 0 && x < m && y >= 0 && y < n;
}
}