題目
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.
解答
題目的意思是說啊，一個二維矩陣，裏面的數值表示的是一棟一棟的大樓的高度，現在要求我們儘可能多的提升這些大樓的高度。要求就是，大樓從上下左右側面看的投影是一樣的。
那麼，也就是說，我們要先找到大樓側面看最高值是多少，因爲矮的樓的投影肯定包含在高樓的投影裏面了。然後遍歷每一個樓，看看要補多少的高度即可。
public int maxIncreaseKeepingSkyline(int[][] grid) {
int[] maxInRow = new int[4];
int[] maxInCol = new int[4];
for (int r = 0; r < grid.length; r++) {
for (int c = 0; c < grid[0].length; c++) {
maxInRow[r] = Math.max(maxInRow[r], grid[r][c]);
maxInCol[c] = Math.max(maxInCol[c], grid[r][c]);
}
}
int x = 0;
for (int r = 0; r < grid.length; r++) {
for (int c = 0; c < grid[0].length; c++) {
x += Math.min(maxInRow[r], maxInCol[c]) - grid[r][c];
}
}
return x;
}