题目
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
题解
方式一：深度优先搜索
public int numIslands(char[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
// 遍历所有元素，从‘1’开始dfs，把这个岛屿全部遍历完
if (grid[i][j] == '1') {
dfs(i, j, grid);
// 结果加一
result++;
}
}
}
return result;
}
public void dfs(int i, int j, char[][] grid) {
if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') {
return;
}
// 遍历当前岛屿时全部把‘1’变为‘0’，避免重复计算
grid[i][j] = '0';
dfs(i - 1, j, grid);
dfs(i + 1, j, grid);
dfs(i, j - 1, grid);
dfs(i, j + 1, grid);
}
方式二：广度优先搜索
public int numIslands(char[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == '1') {
bfs(i, j, grid);
result++;
}
}
}
return result;
}
public void bfs(int i, int j, char[][] grid) {
Deque<Grid> deque = new LinkedList<>();
deque.offer(new Grid(i, j));
grid[i][j] = '0';
while (!deque.isEmpty()) {
Grid g = deque.poll();
int x = g.getX();
int y = g.getY();
if (x - 1 >= 0 && grid[x - 1][y] == '1') {
deque.offer(new Grid(x - 1, y));
grid[x - 1][y] = '0';
}
if (y - 1 >= 0 && grid[x][y - 1] == '1') {
deque.offer(new Grid(x, y - 1));
grid[x][y - 1] = '0';
}
if (x + 1 < grid.length && grid[x + 1][y] == '1') {
deque.offer(new Grid(x + 1, y));
grid[x + 1][y] = '0';
}
if (y + 1 < grid[0].length && grid[x][y + 1] == '1') {
deque.offer(new Grid(x, y + 1));
grid[x][y + 1] = '0';
}
}
}
class Grid {
int x;
int y;
public Grid(int i, int j) {
x = i;
y = j;
}
public int getX() {
return x;
}
public int getY() {
return y;
}
}
总结
算法：DFS、BFS
