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.
思路：碰見(jiàn)是1的點(diǎn)就進(jìn)行寬度搜索乓梨，并把它的所有鄰島標(biāo)記為0馆纳，總島數(shù)+1.
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int res = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
bfs(grid, i, j);
res++;
}
}
}
return res;
}
private void bfs(char[][] grid, int i, int j) {
int[] dx = {0, 1, 0, -1};
int[] dy = {1, 0, -1, 0};
Queue<int[]> q = new LinkedList<>();
int[] coor = {i, j};
q.offer(coor);
while (!q.isEmpty()) {
int[] cur = q.poll();
grid[cur[0]][cur[1]] = '0';
for (int k = 0; k < 4; k++) {
int tx = cur[0] + dx[k];
int ty = cur[1] + dy[k];
if (tx < 0 || tx >= grid.length || ty < 0 || ty >= grid[0].length || grid[tx][ty] == '0') {
continue;
}
int[] neighbor = {tx, ty};
q.offer(neighbor);
}
}
}