Build Post Office

Description

Given a 2D grid, each cell is either an house 1 or empty 0 (the number zero, one), find the place to build a post office, the distance that post office to all the house sum is smallest. Return the smallest distance. Return -1 if it is not possible.

• You can pass through house and empty.
• You only build post office on an empty.
• The distance between house and the post office is Manhattan distance

Example

Example 1:

Input：[[0,1,0,0],[1,0,1,1],[0,1,0,0]]
Output： 6
Explanation：
Placing a post office at (1,1), the distance that post office to all the house sum is smallest.
 
Example 2:

Input：[[0,1,0],[1,0,1],[0,1,0]]
Output： 4
Explanation：
Placing a post office at (1,1), the distance that post office to all the house sum is smallest.
思路：

方法一：

• 由于题中所求两点间距离表示为d(i，j)=|xi-xj|+|yi-yj|，所以可以在x方向和y方向分别计算，然后求和即可，实现一维方向的计算即可。
• 首先遍历网格，为1的方格，将其横纵坐标，即i和j分别存储，然后排序，方便接下来的查找计算。然后利用sumx和sumy前缀和，sumx代表起点到各个点的x方向距离和，sumy同理。
• 再次遍历网格，当前处为0时，首先二分查找该方向i或j的位置，如果有多个相同的，寻找最右边的，即可，设为index。
• 然后求和即可
• sum.get(n) - sum.get(index + 1) - pos * (n - index - 1) + (index + 1) * pos - sum.get(index + 1);
• sum.get(n) - sum.get(index + 1)计算出当前点右侧部分到起点的距离和
• • pos * (n - index - 1) 减去多余部分，因为之前是到起点的距离和，现在要求到当前点的距离和
• (index + 1) * pos - sum.get(index + 1) 此处开始计算当前点左侧部分，index+1个当前点距离和减去index+1到起点的距离和，即为左侧部分的距离和。

方法二：

• 处理前缀：
  prefixSum1记录数组 a 的前缀和，即:prefixSum1[i]=a[0]+a[1]+..+a[i].
  prefixSum2记录数组 prefixSum1 前缀和，prefixSum2即为前 i 个点到第 i 个点的代价和。
• 处理后缀：
  prefixSum1记录数组 a 的后缀和，即:prefixSum1[i]=a[n-1]+a[n-2]+..+a[i].
  prefixSum2记录数组 prefixSum1 的后缀和，prefixSum2即为 i 之后的点到第 i 个点的代价和。

  public class Solution {
      /**
       * @param grid: a 2D grid
       * @return: An integer
       */
     public int shortestDistance(int[][] grid) {
          // Write your code here
          int n = grid.length;
          if (n == 0)
              return -1;
   
          int m = grid[0].length;
          if (m == 0)
              return -1;
   
          List<Integer> sumx = new ArrayList<Integer>();
          List<Integer> sumy = new ArrayList<Integer>();
          List<Integer> x = new ArrayList<Integer>();
          List<Integer> y = new ArrayList<Integer>();
   
          int result = Integer.MAX_VALUE;
          for (int i = 0; i < n; ++i)
              for (int j = 0; j < m; ++j)
                  if (grid[i][j] == 1) {
                      x.add(i);
                      y.add(j);
                  }
   
          Collections.sort(x);
          Collections.sort(y);
   
          int total = x.size();
   
          sumx.add(0);
          sumy.add(0);
          for (int i = 1; i <= total; ++i) {
              sumx.add(sumx.get(i-1) + x.get(i-1));
              sumy.add(sumy.get(i-1) + y.get(i-1));
          }
   
          for (int i = 0; i < n; ++i)
              for (int j = 0; j < m; ++j)
                  if (grid[i][j] == 0) {
                      int cost_x = get_cost(x, sumx, i, total);
                      int cost_y = get_cost(y, sumy, j, total);
                      if (cost_x + cost_y < result)
                          result = cost_x + cost_y;
                  }
   
          return result;
      }
   
      public int get_cost(List<Integer> x, List<Integer> sum, int pos, int n) {
          if (n == 0)
              return 0;
          if (x.get(0) > pos)
              return sum.get(n) - pos * n;
   
          int l = 0, r = n - 1;
          while (l + 1 < r) {
              int mid = l + (r - l) / 2;
              if (x.get(mid) <= pos)
                  l = mid;
              else
                  r = mid - 1;
          }
   
          int index = 0;
          if (x.get(r) <= pos)
              index = r;
          else
              index = l;
          return sum.get(n) - sum.get(index + 1) - pos * (n - index - 1) +
                 (index + 1) * pos - sum.get(index + 1);
      }
  }