You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

  • Each 0 marks an empty land which you can pass by freely.
  • Each 1 marks a building which you cannot pass through.
  • Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1

|   |   |   |   |

0 - 0 - 0 - 0 - 0

|   |   |   |   |

0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

给一个2纬网格，0代表空地可自由通过，1代表建筑物不能通过，2代表障碍物不可通过，找一个位置建房子，使其到所有建筑物的曼哈顿距离之和最小。返回建房子的位置，如果没有这样的位置返回-1。

解法：BFS，对于每一个建筑进行一次BFS计算到每一个可到达的空地的距离，然后对于每一个空地计算到所有建筑的距离和，求出距离和最短的空地。

Java:

public class Solution {

    /**

     * @param grid: the 2D grid

     * @return: the shortest distance

     */

    public int shortestDistance(int[][] grid) {

        if (grid == null || grid.length == 0 || grid[0].length == 0) {

            return 0;

        }

        int m = grid.length, n = grid[0].length;

        int[][] totalDistance = new int[m][n];

        int step = 0, res = 0;

        for (int i = 0; i < m; ++i) {

            for (int j = 0; j < n; ++j) {

                if (grid[i][j] == 1) {

                    res = bfs(grid, i, j, step, totalDistance);

                    step--;

                }

            }

        }

        return res == Integer.MAX_VALUE ? -1 : res;

    }

    private int bfs(int[][] grid, int x, int y, int step, int[][] totalDistance) {

        int res = Integer.MAX_VALUE, m = grid.length, n = grid[0].length;;

        Queue<Integer> queue = new LinkedList<>();

        queue.offer(x * n + y);

        int curDis = 0;

        int[] dirs = {-1, 0, 1, 0, -1};

        while (!queue.isEmpty()) {

            int l = queue.size();

            curDis++;

            while (l-- != 0) {

                int t = queue.poll();

                x = t / n;

                y = t % n;

                for (int i = 0; i < 4; ++i) {

                    int _x = x + dirs[i], _y = y + dirs[i + 1];

                    if (_x >= 0 && _x < m && _y >= 0 && _y < n && grid[_x][_y] == step) {

                        queue.offer(_x * n + _y);

                        totalDistance[_x][_y] += curDis;

                        grid[_x][_y]--;

                        res = Math.min(res, totalDistance[_x][_y]);

                    }

                }

            }

        }

        return res;

    }

}　

Java:

public class Solution {

    /**

     * @param grid: the 2D grid

     * @return: the shortest distance

     */

    int len;

    int m;

    int n;

    int count;

    int sum;

    int[] directions = {0, 1, 0, -1, 0};

    public int shortestDistance(int[][] grid) {

        // write your code here

        m = grid.length;

        n = grid[0].length;

        if (grid == null || m == 0 || n == 0) {

            return -1;

        }

        int house = 0;

        for (int i = 0; i < m; ++i) {

            for (int j = 0; j < n; ++j) {

                if (grid[i][j] == 1) {

                    house++;

                }

            }

        }

        count = 0;

        len = 0;

        sum = 0;

        int minLen = Integer.MAX_VALUE;

        for (int i = 0; i < m; ++i) {

            for (int j = 0; j < n; ++j) {

                if (grid[i][j] == 0) {

                    bfs(grid, i, j);

                    if (count != house) {

                        continue;

                    }else {

                        minLen = Math.min(minLen, sum);

                    }

                }

            }

        }

        return minLen == Integer.MAX_VALUE ? -1: minLen;

    }

    private void bfs(int[][] grid, int i, int j) {

        count = 0;

        len = 0;

        sum = 0;

        Queue<Integer> q = new LinkedList<>();

        Set<Integer> v = new HashSet<>();

        q.offer(i * n + j);

        v.add(i * n + j);

        while (!q.isEmpty()) {

            len++;

            int size = q.size();

            while (size-- != 0) {

                int cur = q.poll();

                int x = cur / n;

                int y = cur % n;

                for (int k = 0; k < 4; ++k) {

                    int nx = x + directions[k];

                    int ny = y + directions[k + 1];

                    if (!v.contains(nx * n + ny) && nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] != 2) {

                        if (grid[nx][ny] == 1) {

                            count++;

                            sum += len;

                            v.add(nx * n + ny);

                            continue;

                        }

                        if (grid[nx][ny] == 0) {

                            q.offer(nx * n + ny);

                            v.add(nx * n + ny);

                        }

                    }

                }

            }

        }

    }

}　　

Python:

# Time:  O(k * m * n), k is the number of the buildings

# Space: O(m * n)

class Solution(object):

    def shortestDistance(self, grid):

        """

        :type grid: List[List[int]]

        :rtype: int

        """

        def bfs(grid, dists, cnts, x, y):

            dist, m, n = 0, len(grid), len(grid[0])

            visited = [[False for _ in xrange(n)] for _ in xrange(m)]

            pre_level = [(x, y)]

            visited[x][y] = True

            while pre_level:

                dist += 1

                cur_level = []

                for i, j in pre_level:

                    for dir in [(-1, 0), (1, 0), (0, -1), (0, 1)]:

                        I, J = i+dir[0], j+dir[1]

                        if 0 <= I < m and 0 <= J < n and grid[I][J] == 0 and not visited[I][J]:

                            cnts[I][J] += 1

                            dists[I][J] += dist

                            cur_level.append((I, J))

                            visited[I][J] = True

                pre_level = cur_level

        m, n, cnt = len(grid),  len(grid[0]), 0

        dists = [[0 for _ in xrange(n)] for _ in xrange(m)]

        cnts = [[0 for _ in xrange(n)] for _ in xrange(m)]

        for i in xrange(m):

            for j in xrange(n):

                if grid[i][j] == 1:

                    cnt += 1

                    bfs(grid, dists, cnts, i, j)

        shortest = float("inf")

        for i in xrange(m):

            for j in xrange(n):

                if dists[i][j] < shortest and cnts[i][j] == cnt:

                    shortest = dists[i][j]

        return shortest if shortest != float("inf") else -1　　

C++:

class Solution {

public:

    int shortestDistance(vector<vector<int>>& grid) {

        int res = INT_MAX, val = 0, m = grid.size(), n = grid[0].size();

        vector<vector<int>> sum = grid;

        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};

        for (int i = 0; i < grid.size(); ++i) {

            for (int j = 0; j < grid[i].size(); ++j) {

                if (grid[i][j] == 1) {

                    res = INT_MAX;

                    vector<vector<int>> dist = grid;

                    queue<pair<int, int>> q;

                    q.push({i, j});

                    while (!q.empty()) {

                        int a = q.front().first, b = q.front().second; q.pop();

                        for (int k = 0; k < dirs.size(); ++k) {

                            int x = a + dirs[k][0], y = b + dirs[k][1];

                            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == val) {

                                --grid[x][y];

                                dist[x][y] = dist[a][b] + 1;

                                sum[x][y] += dist[x][y] - 1;

                                q.push({x, y});

                                res = min(res, sum[x][y]);

                            }

                        }

                    }

                    --val;

                }

            }

        }

        return res == INT_MAX ? -1 : res;

    }

};

C++:

class Solution {

public:

    int shortestDistance(vector<vector<int>>& grid) {

        int res = INT_MAX, buildingCnt = 0, m = grid.size(), n = grid[0].size();

        vector<vector<int>> dist(m, vector<int>(n, 0)), cnt = dist;

        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};

        for (int i = 0; i < m; ++i) {

            for (int j = 0; j < n; ++j) {

                if (grid[i][j] == 1) {

                    ++buildingCnt;

                    queue<pair<int, int>> q;

                    q.push({i, j});

                    vector<vector<bool>> visited(m, vector<bool>(n, false));

                    int level = 1;

                    while (!q.empty()) {

                        int size = q.size();

                        for (int s = 0; s < size; ++s) {

                            int a = q.front().first, b = q.front().second; q.pop();

                            for (int k = 0; k < dirs.size(); ++k) {

                                int x = a + dirs[k][0], y = b + dirs[k][1];

                                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 0 && !visited[x][y]) {

                                    dist[x][y] += level;

                                    ++cnt[x][y];

                                    visited[x][y] = true;

                                    q.push({x, y});

                                }

                            }

                        }

                        ++level;

                    }

                }

            }

        }

        for (int i = 0; i < m; ++i) {

            for (int j = 0; j < n; ++j) {

                if (grid[i][j] == 0 && cnt[i][j] == buildingCnt) {

                    res = min(res, dist[i][j]);

                }

            }

        }

        return res == INT_MAX ? -1 : res;

    }

};

　　

类似题目：

[LeetCode] 286. Walls and Gates 墙和门

[LeetCode] 296. Best Meeting Point 最佳开会地点

　　

All LeetCode Questions List 题目汇总

最新文章

  1. textField和textView限制输入条件
  2. zw版【转发&#183;台湾nvp系列Delphi例程】HALCON HWindowX 02
  3. for循环嵌套的穷举，迭代，以及while的使用和for的转换
  4. iterm2 配色修改
  5. 51nod1265四点共面
  6. heX：用HTML5和Node.JS开发桌面应用
  7. 环境配置与JBoss安装－EJB3.0入门经典学习笔记(1)
  8. python派QQ邮件
  9. GDI绘制时钟效果，与系统时间保持同步，基于Winform
  10. TortoiseSVN文件夹图标不显示
  11. VBS列出windows更新列表
  12. [转]jenkins2 插件安装
  13. 数组之slice,splice,Concact,Reverse,Sort方法
  14. django之缓存的用法, 文件形式与 redis的基本使用
  15. 11g直接路径读、相关参数、10949事件介绍
  16. Windows 10文件夹Shirt+鼠标右键出现“在此处打开命令窗口”
  17. i春秋SRC部落携手同程SRC发布首届漏洞提交任务
  18. Kindle Paper White 使用感受视频上线啦！
  19. Linux下的Make命令实例详解
  20. [.NET开发] C# 合并、拆分PDF文档

热门文章

  1. Django项目中使用plupload插件实现上传图片功能
  2. drf框架总结复习(1）
  3. No Compiler is Provided in this environment Perhaps you are running on JRE rather than a JDK
  4. SpringBoot整合Bubbo
  5. Greenplum常用的gp_toolkit &amp; pg_catalog监控语句
  6. ArrayList 集合：库存管理
  7. 2016android在线测试15-图像 camera2
  8. 「SNOI2019」积木
  9. 微信小程序 Flex局部元素被挤压问题
  10. SpringSecurity的简单入门