LeetCode: Brick Wall

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本文解析了LeetCode上的砖墙问题，通过使用哈希映射记录每列累计宽度及被穿过的行数，实现寻找垂直线以最少穿过砖块的目标。文章提供了详细的算法思路与C++代码实现。

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date: 2018-09-27 04:06:50+00:00
原标题: ‘LeetCode: Brick Wall’
原链接: https://www.dreamoftime0.com/2018/09/27/leetcode-brick-wall/

题目：

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:

<b>Input:</b> 
[[1,2,2,1],
 [3,1,2],
 [1,3,2],
 [2,4],
 [3,1,2],
 [1,3,1,1]]
<b>Output:</b> 2
<b>Explanation:</b> 
<img src="https://leetcode.com/static/images/problemset/brick_wall.png" width="30%"></img>

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.

思路：

建立一个hash map，key是每列到当前为止的和，value为被穿过的行数。

class Solution {
public:
    int leastBricks(vector<vector<int>>& wall) {
        unordered_map<int,int> wallMap;
        int max_crossed=0,cnt_cols=wall.size();
        
        for(int i=0;i<wall.size();++i)
            for(int j=0,sum=0;j<wall[i].size()-1;++j){
                sum+=wall[i][j];
                max_crossed=max(max_crossed,++wallMap[sum]);
            }
        
        return cnt_cols-max_crossed;
        
    }
};

测试:

没有太过需要注意的边界条件。

#include<iostream>
#include<vector>
#include<unordered_map>
using namespace std;

class Solution {
public:
    int leastBricks(vector<vector<int>>& wall) {
        unordered_map<int,int> wallMap;
        int max_crossed=0,cnt_cols=wall.size();

        for(int i=0;i<wall.size();++i)
            for(int j=0,sum=0;j<wall[i].size()-1;++j){//注意不要算上最后一列，因为最后一列是边界，不考虑
                sum+=wall[i][j];
                max_crossed=max(max_crossed,++wallMap[sum]);
            }

        return cnt_cols-max_crossed;

    }
};

#define REQUIRE(sol) {\
    if(sol)\
    cout<<"passed"<<endl;\
    else cout<<"not passed"<<endl;\
}



int main(){
    Solution s;
    vector<vector<int>> wall={{1,2,2,1},{3,1,2},{1,3,2},{2,4},{3,1,2},{1,3,1,1}};
    REQUIRE( s.leastBricks(wall) == 2);
}