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

 

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