leetcode 554. Brick Wall 寻找最多的砖缝 + 使用Map遍历统计

叁歲伎倆 2022-06-03 04:11 102阅读 0赞

There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the least bricks.

The brick wall is represented by a list of rows. Each row is a list of integers representing the width of each brick in this row from left to right.

If your line go through the edge of a brick, then the brick is not considered as crossed. You need to find out how to draw the line to cross the least bricks and return the number of crossed bricks.

You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.

Example:  
Input:  
\[\[1,2,2,1\],  
\[3,1,2\],  
\[1,3,2\],  
\[2,4\],  
\[3,1,2\],  
\[1,3,1,1\]\]  
Output: 2  
Explanation:

![这里写图片描述][SouthEast]

Note:  
The width sum of bricks in different rows are the same and won’t exceed INT\_MAX.  
The number of bricks in each row is in range \[1,10,000\]. The height of wall is in range \[1,10,000\]. Total number of bricks of the wall won’t exceed 20,000.

本题题意很简单，就是找一个缝隙砍墙，求砍破的最少的砖的数量，起初我的想法是直接在每一层砖的缝隙遍历查询统计，后来发现这个想法实在是太蠢了，直接统计缝隙的出现次数即可，这个就是我们要砍的位置，所以直接使用map遍历即可

**这道题其实可以转换为数组求和问题，很棒的想法**

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    
    using namespace std;
    
    
    class Solution
    {
    public:
        int leastBricks(vector<vector<int>>& wall)
        {
            map<int, int> mmp;
            int maxBreak = 0;
            for (vector<int> a : wall)
            {
                int sum = 0;
                for (int i = 0; i<a.size() - 1; i++)
                {
                    sum += a[i];
                    mmp[sum] += 1;
                    maxBreak = max(maxBreak, mmp[sum]);
                }
            }
    
            return wall.size() - maxBreak;
        }
    };

[SouthEast]: /images/20220603/1cddcc44db844e52b3554865dbdf3309.png