LeetCode —— dfs和bfs
797. 所有可能的路径
给你一个有 n 个节点的 有向无环图（DAG），请你找出所有从节点 0 到节点 n-1 的路径并输出（不要求按特定顺序）。 graph[i] 是一个从节点 i 可以访问的所有节点的列表（即从节点 i 到节点 graph[i][j]存在一条有向边）。
示例1：输入：graph = [[1,2],[3],[3],[]] 输出：[[0,1,3],[0,2,3]]
示例2：输入：graph = [[4,3,1],[3,2,4],[3],[4],[]] 输出：[[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]
// dfs
class Solution {
List> list = new ArrayList<>();
LinkedList path = new LinkedList<>();
public List> allPathsSourceTarget(int[][] graph) {
path.add(0);
dfs(graph, 0);
return list;
}
public void dfs(int[][] graph, int node){
if(node == graph.length - 1){
list.add(new ArrayList<>(path));
return;
}
for(int i = 0; i < graph[node].length; i++){
path.add(graph[node][i]);
dfs(graph, graph[node][i]);
path.removeLast();
}
}
} 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。
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
// dfs （以下三种方式等价）
class Solution {
public int numIslands(char[][] grid) {
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == '1'){
count++;
dfs(grid, i, j);
}
}
}
return count;
}
public void dfs(char[][] grid, int i, int j){
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0'){
return;
}
grid[i][j] = '0';
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
}
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public int numIslands(char[][] grid) {
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == '1'){
count++;
dfs(grid, i, j);
}
}
}
return count;
}
public void dfs(char[][] grid, int i, int j){
if(grid[i][j] == '0'){
return;
}
grid[i][j] = '0';
for(int[] pos : positions){
int newRow = i + pos[0];
int newColumn = j + pos[1];
if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length){
continue;
}
dfs(grid, newRow, newColumn);
}
}
}
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
boolean[][] visited; // 通过判断该位置是否被访问过，而不是将1修改为0
public int numIslands(char[][] grid) {
visited = new boolean[grid.length][grid[0].length];
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == '1' && !visited[i][j]){
count++;
dfs(grid, i, j);
}
}
}
return count;
}
public void dfs(char[][] grid, int i, int j){
if(visited[i][j] || grid[i][j] == '0'){
return;
}
visited[i][j] = true;
for(int[] pos : positions){
int newRow = i + pos[0];
int newColumn = j + pos[1];
if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length){
continue;
}
dfs(grid, newRow, newColumn);
}
}
}// bfs 以下三种方式等价
class Solution {
public int numIslands(char[][] grid) {
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == '1'){
count++;
bfs(grid, i, j);
}
}
}
return count;
}
public void bfs(char[][] grid, int i, int j){
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
while(!queue.isEmpty()){
int[] cur = queue.poll();
i = cur[0];
j = cur[1];
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0'){
continue;
}
grid[i][j] = '0';
queue.offer(new int[]{i - 1, j});
queue.offer(new int[]{i + 1, j});
queue.offer(new int[]{i, j - 1});
queue.offer(new int[]{i, j + 1});
}
}
}
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public int numIslands(char[][] grid) {
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == '1'){
count++;
bfs(grid, i, j);
}
}
}
return count;
}
public void bfs(char[][] grid, int i, int j){
Queue queue = new LinkedList();
queue.offer(new int[]{i, j});
grid[i][j] = '0';
while(!queue.isEmpty()){
int[] cur = queue.poll();
for(int[] pos : positions){
int newRow = cur[0] + pos[0];
int newColumn = cur[1] + pos[1];
if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length || grid[newRow][newColumn] == '0'){
continue;
}
queue.offer(new int[]{newRow, newColumn});
grid[newRow][newColumn] = '0';
}
}
}
}
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
boolean[][] visited; // 通过判断该位置是否被访问过，而不是将1修改为0
public int numIslands(char[][] grid) {
visited = new boolean[grid.length][grid[0].length];
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == '1' && !visited[i][j]){
count++;
bfs(grid, i, j);
}
}
}
return count;
}
public void bfs(char[][] grid, int i, int j){
Queue queue = new LinkedList();
queue.offer(new int[]{i, j});
visited[i][j] = true;
while(!queue.isEmpty()){
int[] cur = queue.poll();
for(int[] pos : positions){
int newRow = cur[0] + pos[0];
int newColumn = cur[1] + pos[1];
if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length || grid[newRow][newColumn] == '0' || visited[newRow][newColumn]){
continue;
}
queue.offer(new int[]{newRow, newColumn});
visited[newRow][newColumn] = true;
}
}
}
} 695. 岛屿的最大面积
给你一个大小为 m x n 的二进制矩阵 grid 。岛屿 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在 水平或者竖直的四个方向上 相邻。你可以假设 grid 的四个边缘都被 0（代表水）包围着。岛屿的面积是岛上值为 1 的单元格的数目。计算并返回 grid 中最大的岛屿面积。如果没有岛屿，则返回面积为 0 。
输入：grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]] 输出：6
// dfs
class Solution {
int area;
public int maxAreaOfIsland(int[][] grid) {
int maxArea = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == 1){
area = 0;
dfs(grid, i, j);
maxArea = Math.max(maxArea, area);
}
}
}
return maxArea;
}
public void dfs(int[][] grid, int i, int j){
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
return;
}
area++;
grid[i][j] = 0;
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
}// bfs
class Solution {
int area;
public int maxAreaOfIsland(int[][] grid) {
int maxArea = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == 1){
area = 0;
bfs(grid, i, j);
maxArea = Math.max(maxArea, area);
}
}
}
return maxArea;
}
public void bfs(int[][] grid, int i, int j){
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
while(!queue.isEmpty()){
int[] cur = queue.poll();
i = cur[0];
j = cur[1];
if(i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
area++;
grid[i][j] = 0;
queue.offer(new int[]{i - 1, j});
queue.offer(new int[]{i + 1, j});
queue.offer(new int[]{i, j - 1});
queue.offer(new int[]{i, j + 1});
}
}
}
} 463. 岛屿的周长
给定一个 row x col 的二维网格地图 grid ，其中：grid[i][j] = 1 表示陆地， grid[i][j] = 0 表示水域。网格中的格子 水平和垂直 方向相连（对角线方向不相连）。整个网格被水完全包围，但其中恰好有一个岛屿（或者说，一个或多个表示陆地的格子相连组成的岛屿）。岛屿中没有“湖”（“湖” 指水域在岛屿内部且不和岛屿周围的水相连）。格子是边长为 1 的正方形。网格为长方形，且宽度和高度均不超过 100 。计算这个岛屿的周长。
示例：输入：grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]] 输出：16
// dfs
class Solution {
int count = 0;
public int islandPerimeter(int[][] grid) {
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == 1){
dfs(grid, i, j);
}
}
}
return count;
}
public void dfs(int[][] grid, int i, int j){
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
count++;
return;
}
if(grid[i][j] == 2){
return;
}
grid[i][j] = 2;
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
}
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
boolean[][] visited;
int count;
public int islandPerimeter(int[][] grid) {
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(grid[i][j] == 1 && !visited[i][j]){
dfs(grid, i, j);
}
}
}
return count;
}
public void dfs(int[][] grid, int i, int j){
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
count++;
return;
}
if(visited[i][j] && grid[i][j] == 1){
return;
}
visited[i][j] = true;
for(int[] pos : positions){
int newRow = i + pos[0];
int newColumn = j + pos[1];
dfs(grid, newRow, newColumn);
}
}
}// bfs
class Solution {
int count = 0;
public int islandPerimeter(int[][] grid) {
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);
}
}
}
return count;
}
public void bfs(int[][] grid, int i, int j){
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
while(!queue.isEmpty()){
int[] cur = queue.poll();
i = cur[0];
j = cur[1];
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
count++;
continue;
}
if(grid[i][j] == 2){
continue;
}
grid[i][j] = 2;
queue.offer(new int[]{i - 1, j});
queue.offer(new int[]{i + 1, j});
queue.offer(new int[]{i, j - 1});
queue.offer(new int[]{i, j + 1});
}
}
}
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
boolean[][] visited;
int count = 0;
public int islandPerimeter(int[][] grid) {
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(grid[i][j] == 1 && !visited[i][j]){
bfs(grid, i, j);
}
}
}
return count;
}
public void bfs(int[][] grid, int i, int j){
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
visited[i][j] = true;
while(!queue.isEmpty()){
int[] cur = queue.poll();
for(int[] pos : positions){
int newRow = cur[0] + pos[0];
int newColumn = cur[1] + pos[1];
if(newRow < 0 || newRow >= grid.length || newColumn < 0 || newColumn >= grid[0].length || grid[newRow][newColumn] == 0){
count++;
continue;
}
if(visited[newRow][newColumn] && grid[newRow][newColumn] == 1){
continue;
}
queue.offer(new int[]{newRow, newColumn});
visited[newRow][newColumn] = true;
}
}
}
} 827. 最大人工岛屿
给你一个大小为 n x n 二进制矩阵 grid 。最多 只能将一格 0 变成 1 。返回执行此操作后，grid 中最大的岛屿面积是多少？岛屿 由一组上、下、左、右四个方向相连的 1 形成。
示例：输入: grid = [[1, 0], [0, 1]] 输出: 3
// dfs
class Solution {
int islandSize; // 每个岛屿的大小
public int largestIsland(int[][] grid) {
int[][] positions = new int[][]{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
int m = grid.length;
int n = grid[0].length;
Map map = new HashMap<>(); // key为岛屿编号，value为岛屿大小
int islandNumber = 2; // 岛屿编号从2开始，与0和1进行区分
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(grid[i][j] == 1){
islandSize = 0;
dfs(grid, i, j, islandNumber);
map.put(islandNumber, islandSize);
islandNumber++;
}
}
}
int result = Integer.MIN_VALUE;
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(grid[i][j] != 0){
continue;
}
int curSize = 1;
Set set = new HashSet<>();
for(int[] pos : positions){
int newRow = i + pos[0];
int newColumn = j + pos[1];
if(newRow < 0 || newRow >= m || newColumn < 0 || newColumn >= n || grid[newRow][newColumn] == 0){
continue;
}
islandNumber = grid[newRow][newColumn];
if(set.contains(islandNumber) || !map.containsKey(islandNumber)){
continue;
}
set.add(islandNumber);
curSize += map.get(islandNumber);
}
result = Math.max(result, curSize);
}
}
return result == Integer.MIN_VALUE ? m * n : result;
}
public void dfs(int[][] grid, int i, int j, int islandNumber){
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] != 1){
return;
}
islandSize++;
grid[i][j] = islandNumber;
dfs(grid, i - 1, j, islandNumber);
dfs(grid, i + 1, j, islandNumber);
dfs(grid, i, j - 1, islandNumber);
dfs(grid, i, j + 1, islandNumber);
}
} 1020. 飞地的数量
给你一个大小为 m x n 的二进制矩阵 grid ，其中 0 表示一个海洋单元格、1 表示一个陆地单元格。一次 移动 是指从一个陆地单元格走到另一个相邻（上、下、左、右）的陆地单元格或跨过 grid 的边界。返回网格中 无法 在任意次数的移动中离开网格边界的陆地单元格的数量。
示例：输入：grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]] 输出：3
// dfs
class Solution {
public int numEnclaves(int[][] grid) {
for(int i = 0; i < grid.length; i++){
if(grid[i][0] == 1){ // 左侧
dfs(grid, i, 0);
}
if(grid[i][grid[0].length - 1] == 1){ // 右侧
dfs(grid, i, grid[0].length - 1);
}
}
for(int j = 1; j < grid[0].length - 1; j++){
if(grid[0][j] == 1){ // 上侧
dfs(grid, 0, j);
}
if(grid[grid.length - 1][j] == 1){ // 下侧
dfs(grid, grid.length - 1, j);
}
}
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == 1){
count++;
}
}
}
return count;
}
public void dfs(int[][] grid, int i, int j){
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
return;
}
grid[i][j] = 0;
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
}// bfs
class Solution {
public int numEnclaves(int[][] grid) {
for(int i = 0; i < grid.length; i++){
if(grid[i][0] == 1){ // 左侧
bfs(grid, i, 0);
}
if(grid[i][grid[0].length - 1] == 1){ // 右侧
bfs(grid, i, grid[0].length - 1);
}
}
for(int j = 1; j < grid[0].length - 1; j++){
if(grid[0][j] == 1){ // 上侧
bfs(grid, 0, j);
}
if(grid[grid.length - 1][j] == 1){ // 下侧
bfs(grid, grid.length - 1, j);
}
}
int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
if(grid[i][j] == 1){
count++;
}
}
}
return count;
}
public void bfs(int[][] grid, int i, int j){
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
while(!queue.isEmpty()){
int[] cur = queue.poll();
i = cur[0];
j = cur[1];
if(i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
grid[i][j] = 0;
queue.offer(new int[]{i - 1, j});
queue.offer(new int[]{i + 1, j});
queue.offer(new int[]{i, j - 1});
queue.offer(new int[]{i, j + 1});
}
}
}
} 130. 被围绕的区域
给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。
// dfs
class Solution {
public void solve(char[][] board) {
for(int i = 0; i < board.length; i++){
if(board[i][0] == 'O'){
dfs(board, i, 0);
}
if(board[i][board[0].length - 1] == 'O'){
dfs(board, i, board[0].length - 1);
}
}
for(int j = 1; j < board[0].length - 1; j++){
if(board[0][j] == 'O'){
dfs(board, 0, j);
}
if(board[board.length - 1][j] == 'O'){
dfs(board, board.length - 1, j);
}
}
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == 'A'){
board[i][j] = 'O';
}else if(board[i][j] == 'O'){
board[i][j] = 'X';
}
}
}
}
public void dfs(char[][] board, int i, int j){
if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] == 'X'
|| board[i][j] == 'A'){
return;
}
board[i][j] = 'A';
dfs(board, i - 1, j);
dfs(board, i + 1, j);
dfs(board, i, j - 1);
dfs(board, i, j + 1);
}
}// bfs
class Solution {
public void solve(char[][] board) {
for(int i = 0; i < board.length; i++){
if(board[i][0] == 'O'){
bfs(board, i, 0);
}
if(board[i][board[0].length - 1] == 'O'){
bfs(board, i, board[0].length - 1);
}
}
for(int j = 1; j < board[0].length - 1; j++){
if(board[0][j] == 'O'){
bfs(board, 0, j);
}
if(board[board.length - 1][j] == 'O'){
bfs(board, board.length - 1, j);
}
}
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == 'A'){
board[i][j] = 'O';
}else if(board[i][j] == 'O'){
board[i][j] = 'X';
}
}
}
}
public void bfs(char[][] board, int i, int j){
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
while(!queue.isEmpty()){
int[] cur = queue.poll();
i = cur[0];
j = cur[1];
if(i >= 0 && i < board.length && j >= 0 && j < board[0].length && board[i][j] == 'O'){
board[i][j] = 'A';
queue.offer(new int[]{i - 1, j});
queue.offer(new int[]{i + 1, j});
queue.offer(new int[]{i, j - 1});
queue.offer(new int[]{i, j + 1});
}
}
}
} 417. 太平洋大西洋水流问题
有一个 m × n 的矩形岛屿，与 太平洋 和 大西洋 相邻。 “太平洋” 处于大陆的左边界和上边界，而 “大西洋” 处于大陆的右边界和下边界。这个岛被分割成一个由若干方形单元格组成的网格。给定一个 m x n 的整数矩阵 heights ， heights[r][c] 表示坐标 (r, c) 上单元格 高于海平面的高度 。岛上雨水较多，如果相邻单元格的高度 小于或等于 当前单元格的高度，雨水可以直接向北、南、东、西流向相邻单元格。水可以从海洋附近的任何单元格流入海洋。返回网格标 result 的 2D 列表 ，其中 result[i] = [ri, ci] 表示雨水从单元格 (ri, ci) 流动 既可流向太平洋也可流向大西洋 。
// dfs
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public List> pacificAtlantic(int[][] heights) {
int m = heights.length;
int n = heights[0].length;
boolean[][] pacific = new boolean[m][n];
boolean[][] atlantic = new boolean[m][n];
for(int i = 0; i < m; i++){
dfs(heights, pacific, i, 0);
dfs(heights, atlantic, i, n - 1);
}
for(int j = 0; j < n; j++){
dfs(heights, pacific, 0, j);
dfs(heights, atlantic, m - 1, j);
}
List> list = new ArrayList<>();
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(pacific[i][j] && atlantic[i][j]){
list.add(new ArrayList<>(Arrays.asList(i, j)));
}
}
}
return list;
}
public void dfs(int[][] heights, boolean[][] ocean, int i, int j){
if(ocean[i][j]){
return;
}
ocean[i][j] = true;
for(int[] pos : positions){
int newRow = i + pos[0];
int newColumn = j + pos[1];
if(newRow < 0 || newRow >= heights.length || newColumn < 0 || newColumn >= heights[0].length || heights[i][j] > heights[newRow][newColumn]){
continue;
}
dfs(heights, ocean, newRow, newColumn);
}
}
} // bfs
class Solution {
int[][] positions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public List> pacificAtlantic(int[][] heights) {
int m = heights.length;
int n = heights[0].length;
boolean[][] pacific = new boolean[m][n];
boolean[][] atlantic = new boolean[m][n];
for(int i = 0; i < m; i++){
bfs(heights, pacific, i, 0);
bfs(heights, atlantic, i, n - 1);
}
for(int j = 0; j < n; j++){
bfs(heights, pacific, 0, j);
bfs(heights, atlantic, m - 1, j);
}
List> list = new ArrayList<>();
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(pacific[i][j] && atlantic[i][j]){
list.add(new ArrayList<>(Arrays.asList(i, j)));
}
}
}
return list;
}
public void bfs(int[][] heights, boolean[][] ocean, int i, int j){
ocean[i][j] = true;
Queue queue = new LinkedList<>();
queue.offer(new int[]{i, j});
while(!queue.isEmpty()){
int[] cur = queue.poll();
for(int[] pos : positions){
int newRow = cur[0] + pos[0];
int newColumn = cur[1] + pos[1];
if(newRow < 0 || newRow >= heights.length || newColumn < 0 || newColumn >= heights[0].length || heights[cur[0]][cur[1]] > heights[newRow][newColumn] || ocean[newRow][newColumn]){
continue;
}
ocean[newRow][newColumn] = true;
queue.offer(new int[]{newRow, newColumn});
}
}
}
} 841. 钥匙和房间
有 n 个房间，房间按从 0 到 n - 1 编号。最初，除 0 号房间外的其余所有房间都被锁住。你的目标是进入所有的房间。然而，你不能在没有获得钥匙的时候进入锁住的房间。当你进入一个房间，你可能会在里面找到一套不同的钥匙，每把钥匙上都有对应的房间号，即表示钥匙可以打开的房间。你可以拿上所有钥匙去解锁其他房间。给你一个数组 rooms 其中 rooms[i] 是你进入 i 号房间可以获得的钥匙集合。如果能进入 所有 房间返回 true，否则返回 false。
示例1：输入：rooms = [[1],[2],[3],[]] 输出：true
示例2：输入：rooms = [[1,3],[3,0,1],[2],[0]] 输出：false
// dfs
class Solution {
public boolean canVisitAllRooms(List> rooms) {
boolean[] visited = new boolean[rooms.size()];
dfs(rooms, visited, 0);
for(int i = 0; i < visited.length; i++){
if(!visited[i]){
return false;
}
}
return true;
}
public void dfs(List> rooms, boolean[] visited, int key){
if(visited[key]){
return;
}
visited[key] = true;
List curRoom = rooms.get(key);
for(int i = 0; i < curRoom.size(); i++){
dfs(rooms, visited, curRoom.get(i));
}
}
} // bfs
class Solution {
public boolean canVisitAllRooms(List> rooms) {
boolean[] visited = new boolean[rooms.size()];
bfs(rooms, visited);
for(int i = 0; i < visited.length; i++){
if(!visited[i]){
return false;
}
}
return true;
}
public void bfs(List> rooms, boolean[] visited){
Queue queue = new LinkedList<>();
queue.offer(0);
while(!queue.isEmpty()){
int curKey = queue.poll();
visited[curKey] = true;
List curRoom = rooms.get(curKey);
for(int i = 0; i < curRoom.size(); i++){
if(visited[curRoom.get(i)]){
continue;
}
queue.offer(curRoom.get(i));
}
}
}
} 127. 单词接龙
字典 wordList 中从单词 beginWord 和 endWord 的 转换序列 是一个按下述规格形成的序列 beginWord -> s1 -> s2 -> ... -> sk：①每一对相邻的单词只差一个字母；②对于 1 <= i <= k 时，每个 si 都在 wordList 中。注意， beginWord 不需要在 wordList 中；③sk == endWord。给你两个单词 beginWord 和 endWord 和一个字典 wordList ，返回 从 beginWord 到 endWord 的 最短转换序列 中的 单词数目 。如果不存在这样的转换序列，返回 0 。
示例：输入：beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] 输出：5；
解释：一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
class Solution {
public int ladderLength(String beginWord, String endWord, List wordList) {
Set wordSet = new HashSet<>(wordList); // 转换为HashSet, 加快速度
if(!wordSet.contains(endWord)){
return 0;
}
// bfs 广搜只要搜到了终点，那么一定是最短的路径
Queue queue = new LinkedList<>();
queue.offer(beginWord);
Map map = new HashMap<>(); // 存储单词对应的路径长度
map.put(beginWord, 1);
while(!queue.isEmpty()){
String word = queue.poll();
int pathLen = map.get(word);
for(int i = 0; i < word.length(); i++){
char[] chars = word.toCharArray();
for(char j = 'a'; j <= 'z'; j++){
chars[i] = j;
String newWord = String.valueOf(chars);
if(newWord.equals(endWord)){
return pathLen + 1;
}
if(wordSet.contains(newWord) && !map.containsKey(newWord)){
queue.offer(newWord);
map.put(newWord, pathLen + 1);
}
}
}
}
return 0;
}
}