• LC 807. Max Increase to Keep City Skyline


    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.

    Runtime: 8 ms, faster than 49.01% of C++ online submissions for Max Increase to Keep City Skyline.

    保存横竖最大值的最小值,减去当前值即可。

    #include <algorithm>
    class Solution {
    public:
      int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
        int r = grid.size();
        int c = grid[0].size();
        vector<int> rgrid(r,0);
        vector<int> cgrid(c,0);
        for(int i=0; i<r; i++) rgrid[i] = *max_element(grid[i].begin(), grid[i].end());
        for(int i=0; i<c; i++){
          int maxval = 0;
          for(int j=0; j<r; j++){
            maxval = max(maxval, grid[j][i]);
          }
          cgrid[i] = maxval;
        }
        //for(auto x : cgrid) cout << x << endl;
        //for(auto x : rgrid) cout << x << endl;
        int ret = 0, tmp = 0;
        for(int i=0; i<r; i++){
          for(int j=0; j<c; j++){
            tmp = min(rgrid[i],cgrid[j]) - grid[i][j];
            //cout << tmp << " " << grid[i][j] << endl;
            ret += tmp > 0 ? tmp : 0;
          }
        }
        return ret;
      }
    };
  • 相关阅读:
    C#多线程操作界面控件的解决方案
    InvokeHelper,让跨线程访问/修改主界面控件不再麻烦
    .netCF中后台多线程与UI界面交互的冻结问题
    c#设计模式第一天
    C#代理
    界面
    第一章面向对象涉及原则
    C# 为webBrowser设置代理
    设计模式等
    下载: Intel® 64 and IA32 Architectures Software Developer Manuals
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10177235.html
Copyright © 2020-2023  润新知