LeetCode 807 Max Increase to Keep City Skyline

LeetCode 807 Max Increase to Keep City Skyline

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題目分析

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.

計算每棟樓最多增高多少，不會該變每行每列最大值

思路

獲取每行每列最大值分別存放，遍歷矩陣時再計算當前樓最高可爲所在行列較小值，計算差值求和即可，如果使用python的numpy估計更舒服了。

代碼實現

class Solution {
    void update(int &a, int b) {
        a = max(a, b);
    }
public:
    int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        if (grid.size() == 0) {
            return 0;
        }
        int sum = 0;
        int limit = grid.size();
        vector<int> row_max(limit, 0);
        vector<int> col_max(limit, 0);
        // 獲取行列最大值
        for (int i = 0; i < limit; ++i) {
            for (int j = 0; j < limit; ++j) {
                update(row_max[i], grid[i][j]);
            }
        }              
        for (int i = 0; i < limit; ++i) {
            for (int j = 0; j < limit; ++j) {
                update(col_max[i], grid[j][i]);
            }
        }
        // 計算每個樓增高的數量並求和
        for (int i = 0; i < limit; ++i) {
            for (int j = 0; j < limit; ++j) {
                sum += min(row_max[i], col_max[j]) - grid[i][j];
            }
        }

        return sum;
    }
};

感想

不知道有沒有什麼簡單一點的算法，總覺得這個獲取行和列最大值的兩個循環可以短一點的，嗯？