[Leetcode] 554. Brick Wall 解题报告

最新推荐文章于 2021-08-07 09:29:53 发布

原创最新推荐文章于 2021-08-07 09:29:53 发布 · 392 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #Hash Table

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

本文介绍了一种算法，用于确定在给定的砖墙布局中如何绘制一条垂直线以穿越最少数量的砖块。通过分析每行砖块的宽度并利用哈希表记录砖块接头的位置，该算法能够找到最优的切割位置。

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:

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

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：

用一个哈希表记录每个横坐标处砖头接头出现的次数。最后找出出现砖头接头最多的横坐标，即为需要切割的地方。当然我们用二叉树也可以实现。

代码：

class Solution {
public:
    int leastBricks(vector<vector<int>>& wall) {
        unordered_map<int, int> hash;       // {x, num}
        for (int r = 0; r < wall.size(); ++r) {
            int sum = 0;
            for (int c = 0; c + 1 < wall[r].size(); ++c) {
                sum += wall[r][c];
                ++hash[sum];
            }
        }
        int no_cross = 0;
        for (auto it = hash.begin(); it != hash.end(); ++it) {
            no_cross = max(no_cross, it->second);
        }
        return wall.size() - no_cross;
    }
};