807. Max Increase to Keep City Skyline

墨蓝 2022-05-25 04:54 119阅读 0赞

/\*  
problom description  
  
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.  
  
  
\*/  
/\*  
  
the method of solution:  
calculate the sum of the prime matrix , initialize the prime array to store the max data of every row and column , then compare the max data of every row to the max data of every column,if data of row is bigger ,data of matrix after change is data of row,or the data of matrix after change is data of column  
 my method beats 92%  
\*/  
class Solution \{  
    public int maxIncreaseKeepingSkyline(int\[\]\[\] grid) \{  
        int total=0,total\_2=0,i=0,j=0;  
        int n = grid.length;  
        int m = grid\[0\].length;  
        int\[\] ud=new int\[51\];  
        int\[\] lr=new int\[51\];  
        int max=0;  
          
        for(i=0;i<n;i++)  
            for(j=0;j<m;j++)  
                total+=grid\[i\]\[j\];  
          
        for(i=0;i<n;i++)\{  
            max = grid\[i\]\[0\];  
            for(j=0;j<m;j++)\{  
                if(max<grid\[i\]\[j\])  
                    max = grid\[i\]\[j\];  
            \}  
            lr\[i\] = max;  
        \}  
  
        for(j=0;j<m;j++)\{  
            max = grid\[0\]\[j\];  
            for(i=0;i<n;i++)\{  
                if(max<grid\[i\]\[j\])  
                    max = grid\[i\]\[j\];  
            \}  
            ud\[j\] = max;  
        \}  
  
        for(i=0;i<n;i++)\{  
            for(j=0;j<m;j++)\{  
                if(lr\[i\]<ud\[j\])\{  
                    grid\[i\]\[j\] = lr\[i\];  
                \}  
                else\{  
                    grid\[i\]\[j\] = ud\[j\];  
                \}  
            \}  
        \}  
          
        for(i=0;i<n;i++)  
            for(j=0;j<m;j++)  
                total\_2+=grid\[i\]\[j\];  
  
        return total\_2-total;  
    \}  
\}