[leetcode] 554. Brick Wall

原创

是念 2022-08-11 17:31:09 博主文章分类：C++ ©著作权

文章标签 sed 参考文献 hash表 文章分类 OpenStack 云计算

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Description

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:

[leetcode] 554. Brick Wall_sed

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.

分析

题目的意思是：
现在有一面有很多砖堆砌的墙，砖的大小不一，现在有从垂直切一倒，求在哪里切只需要破坏最少的砖

我们用hash表建立断点与频率的映射，所以我们只要找到出现断点频率最多的地方断开就能损坏最小的砖。

代码

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