200. 岛屿数量
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。
提示：
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 300
- grid[i][j] 的值为 '0' 或 '1'
思路
注意题目中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。
也就是说斜角度链接是不算了， 例如示例二，是三个岛屿，如图：
这道题题目是 DFS，BFS，并查集，基础题目。
本题思路，是用遇到一个没有遍历过的节点陆地，计数器就加一，然后把该节点陆地所能遍历到的陆地都标记上。
在遇到标记过的陆地节点和海洋节点的时候直接跳过。 这样计数器就是最终岛屿的数量。
那么如果把节点陆地所能遍历到的陆地都标记上呢，就可以使用 DFS，BFS或者并查集。
广度优先搜索
不少同学用广搜做这道题目的时候，超时了。 这里有一个广搜中很重要的细节：
根本原因是只要 加入队列就代表 走过，就需要标记，而不是从队列拿出来的时候再去标记走过。
很多同学可能感觉这有区别吗？
如果从队列拿出节点，再去标记这个节点走过，就会发生下图所示的结果，会导致很多节点重复加入队列。
超时写法 （从队列中取出节点再标记）
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
queue<pair<int, int>> que;
que.push({x, y});
while(!que.empty()) {
pair<int ,int> cur = que.front(); que.pop();
int curx = cur.first;
int cury = cur.second;
visited[curx][cury] = true; // 从队列中取出在标记走过
for (int i = 0; i < 4; i++) {
int nextx = curx + dir[i][0];
int nexty = cury + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') {
que.push({nextx, nexty});
}
}
}
}
加入队列 就代表走过，立刻标记，正确写法：
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
queue<pair<int, int>> que;
que.push({x, y});
visited[x][y] = true; // 只要加入队列，立刻标记
while(!que.empty()) {
pair<int ,int> cur = que.front(); que.pop();
int curx = cur.first;
int cury = cur.second;
for (int i = 0; i < 4; i++) {
int nextx = curx + dir[i][0];
int nexty = cury + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') {
que.push({nextx, nexty});
visited[nextx][nexty] = true; // 只要加入队列立刻标记
}
}
}
}
以上两个版本其实，其实只有细微区别，就是 visited[x][y] = true; 放在的地方，着去取决于我们对 代码中队列的定义，队列中的节点就表示已经走过的节点。 所以只要加入队列，立即标记该节点走过。
本题完整广搜代码：
class Solution {
private:
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
queue<pair<int, int>> que;
que.push({x, y});
visited[x][y] = true; // 只要加入队列，立刻标记
while(!que.empty()) {
pair<int ,int> cur = que.front(); que.pop();
int curx = cur.first;
int cury = cur.second;
for (int i = 0; i < 4; i++) {
int nextx = curx + dir[i][0];
int nexty = cury + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') {
que.push({nextx, nexty});
visited[nextx][nexty] = true; // 只要加入队列立刻标记
}
}
}
}
public:
int numIslands(vector<vector<char>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
result++; // 遇到没访问过的陆地，+1
bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
}
}
}
return result;
}
};
其他语言版本
Java
class Solution {
boolean[][] visited;
int[][] move = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
public int numIslands(char[][] grid) {
int res = 0;
visited = new boolean[grid.length][grid[0].length];
for(int i = 0; i < grid.length; i++) {
for(int j = 0; j < grid[0].length; j++) {
if(!visited[i][j] && grid[i][j] == '1') {
bfs(grid, i, j);
res++;
}
}
}
return res;
}
//将这片岛屿上的所有陆地都访问到
public void bfs(char[][] grid, int y, int x) {
Deque<int[]> queue = new ArrayDeque<>();
queue.offer(new int[]{y, x});
visited[y][x] = true;
while(!queue.isEmpty()) {
int[] cur = queue.poll();
int m = cur[0];
int n = cur[1];
for(int i = 0; i < 4; i++) {
int nexty = m + move[i][0];
int nextx = n + move[i][1];
if(nextx < 0 || nexty == grid.length || nexty < 0 || nextx == grid[0].length) continue;
if(!visited[nexty][nextx] && grid[nexty][nextx] == '1') {
queue.offer(new int[]{nexty, nextx});
visited[nexty][nextx] = true; //只要加入队列就标记为访问
}
}
}
}
}
Python
BFS solution
class Solution:
def __init__(self):
self.dirs = [[0, 1], [1, 0], [-1, 0], [0, -1]]
def numIslands(self, grid: List[List[str]]) -> int:
m = len(grid)
n = len(grid[0])
visited = [[False]*n for _ in range(m)]
res = 0
for i in range(m):
for j in range(n):
if visited[i][j] == False and grid[i][j] == '1':
res += 1
self.bfs(grid, i, j, visited) # Call bfs within this condition
return res
def bfs(self, grid, i, j, visited):
q = deque()
q.append((i,j))
visited[i][j] = True
while q:
x, y = q.popleft()
for k in range(4):
next_i = x + self.dirs[k][0]
next_j = y + self.dirs[k][1]
if next_i < 0 or next_i >= len(grid):
continue
if next_j < 0 or next_j >= len(grid[0]):
continue
if visited[next_i][next_j]:
continue
if grid[next_i][next_j] == '0':
continue
q.append((next_i, next_j))
visited[next_i][next_j] = True
JavaScript
var numIslands = function (grid) {
let dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
let bfs = (grid, visited, x, y) => {
let queue = [];
queue.push([x, y]);
visited[x][y] = true;
while (queue.length) {
let top = queue.shift();//取出队列头部元素
console.log(top)
for (let i = 0; i < 4; i++) {
let nextX = top[0] + dir[i][0]
let nextY = top[1] + dir[i][1]
if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length)
continue;
if (!visited[nextX][nextY] && grid[nextX][nextY] === "1") {
queue.push([nextX, nextY])
visited[nextX][nextY] = true
}
}
}
}
let visited = new Array(grid.length).fill().map(() => Array(grid[0].length).fill(false))
let res = 0
for (let i = 0; i < grid.length; i++) {
for (let j = 0; j < grid[i].length; j++) {
if (!visited[i][j] && grid[i][j] === "1") {
++res;
bfs(grid, visited, i, j);
}
}
}
return res
};
TypeScript
function numIslands2(grid: string[][]): number {
// 四个方向
const dir: number[][] = [[0, 1], [1, 0], [-1, 0], [0, -1]];
const [m, n]: [number, number] = [grid.length, grid[0].length];
function dfs(grid: string[][], visited: boolean[][], x: number, y: number) {
const queue: number[][] = [[x, y]];
while (queue.length !== 0) {
//取出队列头部元素
const top: number[] = queue.shift()!;
for (let i = 0; i < 4; i++) {
const nextX: number = top[0] + dir[i][0];
const nextY: number = top[1] + dir[i][1];
// 越界了，直接跳过
if (nextX < 0 || nextX >= m || nextY < 0 || nextY >= n) {
continue;
}
if (!visited[nextX][nextY] && grid[nextX][nextY] === '1') {
queue.push([nextX, nextY]);
// 只要加入队列立刻标记
visited[nextX][nextY] = true;
}
}
}
}
const visited: boolean[][] = Array.from({ length: m }, _ => new Array(n).fill(false));
let result = 0;
for (let i = 0; i < m; i++) {
for (let k = 0; k < n; k++) {
if (!visited[i][k] && grid[i][k] === '1') {
++result; // 遇到没访问过的陆地，+1
visited[i][k] = true;
dfs(grid, visited, i, k); // 将与其链接的陆地都标记上 true
}
}
}
return result;
}
Go
var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
func numIslands(grid [][]byte) int {
res := 0
visited := make([][]bool, len(grid))
for i := 0; i < len(grid); i++ {
visited[i] = make([]bool, len(grid[0]))
}
for i, rows := range grid {
for j, v := range rows {
if v == '1' && !visited[i][j] {
res++
bfs(grid, visited, i, j)
}
}
}
return res
}
func bfs(grid [][]byte, visited [][]bool, i, j int) {
queue := [][2]int{{i, j}}
visited[i][j] = true // 标记已访问，循环中标记会导致重复
for len(queue) > 0 {
cur := queue[0]
queue = queue[1:]
for _, d := range DIRECTIONS {
x, y := cur[0]+d[0], cur[1]+d[1]
if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) {
continue
}
if grid[x][y] == '1' && !visited[x][y] {
visited[x][y] = true
queue = append(queue, [2]int{x, y})
}
}
}
}
Rust
use std::collections::VecDeque;
impl Solution {
const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];
pub fn num_islands(grid: Vec<Vec<char>>) -> i32 {
let mut visited = vec![vec![false; grid[0].len()]; grid.len()];
let mut res = 0;
for (i, chars) in grid.iter().enumerate() {
for (j, &c) in chars.iter().enumerate() {
if !visited[i][j] && c == '1' {
res += 1;
Self::bfs(&grid, &mut visited, (i as i32, j as i32));
}
}
}
res
}
pub fn bfs(grid: &Vec<Vec<char>>, visited: &mut Vec<Vec<bool>>, (x, y): (i32, i32)) {
let mut queue = VecDeque::new();
queue.push_back((x, y));
visited[x as usize][y as usize] = true;
while let Some((cur_x, cur_y)) = queue.pop_front() {
for (dx, dy) in Self::DIRECTIONS {
let (nx, ny) = (cur_x + dx, cur_y + dy);
if nx < 0 || nx >= grid.len() as i32 || ny < 0 || ny >= grid[0].len() as i32 {
continue;
}
let (nx, ny) = (nx as usize, ny as usize);
if grid[nx][ny] == '1' && !visited[nx][ny] {
visited[nx][ny] = true;
queue.push_back((nx as i32, ny as i32));
}
}
}
}
}