807. Max Increase to Keep City Skyline

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题目描述：

In a 2 dimensional array grid, each value grid[i][j] represents the height of a building located there. We are allowed to increase the height of any number of buildings, by any amount (the amounts can be different for different buildings). Height 0 is considered to be a building as well. 

At the end, the "skyline" when viewed from all four directions of the grid, i.e. top, bottom, left, and right, must be the same as the skyline of the original grid. A city‘s skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. See the following example.

What is the maximum total sum that the height of the buildings can be increased?

Example:
Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation: 
The grid is:
[ [3, 0, 8, 4], 
  [2, 4, 5, 7],
  [9, 2, 6, 3],
  [0, 3, 1, 0] ]

The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]

The grid after increasing the height of buildings without affecting skylines is:

gridNew = [ [8, 4, 8, 7],
            [7, 4, 7, 7],
            [9, 4, 8, 7],
            [3, 3, 3, 3] ]

Notes:

• 1 < grid.length = grid[0].length <= 50.
• All heights grid[i][j] are in the range [0, 100].
• All buildings in grid[i][j] occupy the entire grid cell: that is, they are a 1 x 1 x grid[i][j] rectangular prism.

解题思路：

先得到每一行所对应最高建筑的高度row[i]和每一列对应的最高建筑的高度col[j]。要增加建筑面积，但是为了不改变横向和纵向看去城市最高的建筑高度，

对于每一个建筑[i][j]：

　　(1)是其所在行或列的最高建筑时，不能改变其高度。

　　(2)其不是所在行或列的最高建筑时，可以增加高度至col[j]和row[i]中最小值。

代码：

 1 int maxIncreaseKeepingSkyline(int** grid, int gridRowSize, int *gridColSizes) {
 2     int* row = (int*) malloc(sizeof(int) * gridRowSize);
 3     int* col = (int*) malloc(sizeof(int) * gridColSizes[0]);
 4     for (int i = 0; i < gridRowSize; ++i) {
 5         int max = 0;
 6         for (int j = 0; j < gridColSizes[0]; ++j) {
 7             if (grid[i][j] > max)
 8                 max = grid[i][j];
 9         }
10         row[i] = max;
11     }
12     for (int i = 0; i < gridColSizes[0]; ++i) {
13         int max = 0;
14         for (int j = 0; j < gridRowSize; ++j) {
15             if (grid[j][i] > max)
16                 max = grid[j][i];
17         }
18         col[i] = max;
19     }
20     int added = 0;
21     for (int i = 0; i < gridRowSize; ++i) {
22         for (int j = 0; j < gridColSizes[0]; ++j) {
23             int max = row[i] < col[j] ? row[i] : col[j];
24             if (grid[i][j] < max)
25                 added += max - grid[i][j];
26         }
27     }
28     return added;
29 }

807. Max Increase to Keep City Skyline

标签：locate   ons   tin   NPU   red   oca   nts   otto   Plan   

原文地址：https://www.cnblogs.com/gsz-/p/9379704.html