在某区域中，找出连续空闲区域的最大范围
题目描述
看电影选座位，座位用m*n的二维矩阵表示，用1代表该作为已选，0代表未选。求选择出最大的相邻空座位个数（上下左右为相邻）。
示例一：
- 输入：
4
7
1 0 0 1 0 0 0
1 0 0 0 0 1 1
0 0 0 1 0 0 0
1 1 0 1 1 0 0
- 输出：
18
示例二：
- 输入：
4
7
1 0 0 1 0 0 0
1 0 0 1 0 1 1
0 0 0 1 0 0 0
1 1 0 1 1 0 0
- 输出：
9
解题思路
1. 递归
查找所有未访问过的坐标点，从某一点开始，递归遍历某点的上下左右，直到所有坐标点都访问过。
class Solution {
public int maxContinuousVacantSeatsNum(int[][] nums, int m, int n) {
if (m <= 0 || n <= 0) {
return 0;
}
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (nums[i][j] == 1) {
visited[i][j] = true;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
// 访问过的坐标点直接跳过
if (visited[i][j]) {
continue;
}
int res = dfs(nums, m, n, i, j, visited);
ans = Math.max(ans, res);
}
}
return ans;
}
public int dfs(int[][] nums, int m, int n, int startX, int startY, boolean[][] visited) {
if (startX >= m || startY >= n || startX < 0 || startY < 0 || visited[startX][startY]) {
return 0;
}
visited[startX][startY] = true;
return 1 + dfs(nums, m, n, startX + 1, startY, visited) + dfs(nums, m, n, startX - 1, startY, visited) + dfs(nums, m, n, startX, startY + 1, visited) + dfs(nums, m, n, startX, startY - 1, visited);
}
}
2. BFS
使用队列记录压缩的二维坐标，这里注意矩阵坐标的对称性，要分为上、下三角矩阵来存储坐标信息。
class Solution {
public int maxContinuousVacantSeatsNum(int[][] nums, int m, int n) {
if (m <= 0 || n <= 0) {
return 0;
}
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (nums[i][j] == 1) {
visited[i][j] = true;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (visited[i][j]) {
continue;
}
int res = bfs(nums, m, n, i, j, visited);
ans = Math.max(ans, res);
}
}
return ans;
}
public int bfs(int[][] nums, int m, int n, int startX, int startY, boolean[][] visited) {
// 主对角线下三角矩阵队列
Queue<Integer> queue1 = new LinkedList<>();
// 主对角线上三角矩阵队列
Queue<Integer> queue2 = new LinkedList<>();
int count = 0;
if (startX >= startY) {
queue1.offer(startX * n + startY);
} else {
queue2.offer(startX * n + startY);
}
while (!queue1.isEmpty() || !queue2.isEmpty()) {
int dot = !queue1.isEmpty() ? queue1.poll() : queue2.poll();
// 上三角矩阵和下三角矩阵，关于主对角线对称的部分，其压缩计算后的值相同，因此需要根据row和col的大小分别计算存取
int row = dot / n, col = dot % n;
if (row >= m || col >= n) {
continue;
}
if (!visited[row][col]) {
visited[row][col] = true;
if (row >= col) {
queue1.offer(row * n + col);
} else {
queue2.offer(row * n + col);
}
count++;
}
if (col < n - 1 && !visited[row][col + 1]) {
visited[row][col + 1] = true;
if (row >= col + 1) {
queue1.offer(row * n + col + 1);
} else {
queue2.offer(row * n + col + 1);
}
count++;
}
if (col > 0 && !visited[row][col - 1]) {
visited[row][col - 1] = true;
if (row >= col - 1) {
queue1.offer(row * n + col - 1);
} else {
queue2.offer(row * n + col - 1);
}
count++;
}
if (row < m - 1 && !visited[row + 1][col]) {
visited[row + 1][col] = true;
if (row + 1 >= col) {
queue1.offer((row + 1) * n + col);
} else {
queue2.offer((row + 1) * n + col);
}
count++;
}
if (row > 0 && !visited[row - 1][col]) {
visited[row - 1][col] = true;
if (row - 1 >= col) {
queue1.offer((row - 1) * n + col);
} else {
queue2.offer((row - 1) * n + col);
}
count++;
}
}
return count;
}
}
全部代码
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
// 输入次数
int k = scanner.nextInt();
for (int c = 0; c < k; c++) {
int m = scanner.nextInt(), n = scanner.nextInt();
int[][] nums = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
nums[i][j] = scanner.nextInt();
}
}
Main main = new Main();
System.out.println(main.maxContinuousVacantSeatsNum(nums, m, n));
}
}
private int ans = Integer.MIN_VALUE;
public int maxContinuousVacantSeatsNum(int[][] nums, int m, int n) {
if (m <= 0 || n <= 0) {
return 0;
}
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (nums[i][j] == 1) {
visited[i][j] = true;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (visited[i][j]) {
continue;
}
int res = dfs(nums, m, n, i, j, visited);
ans = Math.max(ans, res);
}
}
return ans;
}
public int dfs(int[][] nums, int m, int n, int startX, int startY, boolean[][] visited) {
if (startX >= m || startY >= n || startX < 0 || startY < 0 || visited[startX][startY]) {
return 0;
}
visited[startX][startY] = true;
return 1 + dfs(nums, m, n, startX + 1, startY, visited) + dfs(nums, m, n, startX - 1, startY, visited) + dfs(nums, m, n, startX, startY + 1, visited) + dfs(nums, m, n, startX, startY - 1, visited);
}
public int bfs(int[][] nums, int m, int n, int startX, int startY, boolean[][] visited) {
// 主对角线下三角矩阵队列
Queue<Integer> queue1 = new LinkedList<>();
// 主对角线上三角矩阵队列
Queue<Integer> queue2 = new LinkedList<>();
int count = 0;
if (startX >= startY) {
queue1.offer(startX * n + startY);
} else {
queue2.offer(startX * n + startY);
}
while (!queue1.isEmpty() || !queue2.isEmpty()) {
int dot = !queue1.isEmpty() ? queue1.poll() : queue2.poll();
// 上三角矩阵和下三角矩阵，关于主对角线对称的部分，其压缩计算后的值相同，因此需要根据row和col的大小分别计算存取
int row = dot / n, col = dot % n;
if (row >= m || col >= n) {
continue;
}
if (!visited[row][col]) {
visited[row][col] = true;
if (row >= col) {
queue1.offer(row * n + col);
} else {
queue2.offer(row * n + col);
}
count++;
}
if (col < n - 1 && !visited[row][col + 1]) {
visited[row][col + 1] = true;
if (row >= col + 1) {
queue1.offer(row * n + col + 1);
} else {
queue2.offer(row * n + col + 1);
}
count++;
}
if (col > 0 && !visited[row][col - 1]) {
visited[row][col - 1] = true;
if (row >= col - 1) {
queue1.offer(row * n + col - 1);
} else {
queue2.offer(row * n + col - 1);
}
count++;
}
if (row < m - 1 && !visited[row + 1][col]) {
visited[row + 1][col] = true;
if (row + 1 >= col) {
queue1.offer((row + 1) * n + col);
} else {
queue2.offer((row + 1) * n + col);
}
count++;
}
if (row > 0 && !visited[row - 1][col]) {
visited[row - 1][col] = true;
if (row - 1 >= col) {
queue1.offer((row - 1) * n + col);
} else {
queue2.offer((row - 1) * n + col);
}
count++;
}
}
return count;
}
}
类似题目
剑指 Offer II 105. 岛屿的最大面积
给定一个由 0 和 1 组成的非空二维数组 grid ，用来表示海洋岛屿地图。
一个 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。
找到给定的二维数组中最大的岛屿面积。如果没有岛屿，则返回面积为 0 。
class Solution {
private int ans = 0;
public int maxAreaOfIsland(int[][] grid) {
int m = grid.length, n = grid[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0) continue;
ans = Math.max(ans, dfs(grid, i, j, m, n));
}
}
return ans;
}
public int dfs(int[][] grid, int startX, int startY, int m, int n) {
if (startX < 0 || startY < 0 || startX >= m || startY >= n || grid[startX][startY] == 0) {
return 0;
}
grid[startX][startY] = 0;
return 1 + dfs(grid, startX + 1, startY, m, n) + dfs(grid, startX - 1, startY, m, n) + dfs(grid, startX, startY + 1, m, n) + dfs(grid, startX, startY - 1, m, n);
}
}
