554. Brick Wall

问题描述：

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:

1. The width sum of bricks in different rows are the same and won't exceed INT_MAX.
2. 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.

解题思路：

这道题要我们求穿过的墙最少，那么反方向想一下，就是穿过的缝隙最多。

我们可以遍历这面墙，只对每块砖的边界进行操作，注意不包括最后一块砖（因为两边的肯定都是0）

可以用map存储缝隙的位置与行数。

找到最大的个数，用行数减去缝隙的行数，就是穿过的砖数啦

代码：

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