[LeetCode] 317. Shortest Distance from All Buildings 建筑物的最短距离
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; } };
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