[leetcode] 554. Brick Wall

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.

这道题是计算穿墙“打洞”的最小代价，题目难度为Medium。

要凿穿最少的砖块，就要找同一纵向线上有最多砖缝的位置。在同一纵向线上的砖缝，其左侧砖块的总长度是相同的，我们可以拿砖缝左侧砖块的总长度来标记每个砖缝，这样遍历每一行的砖块将所有砖缝位置计数后存入Hash Table中，最终遍历Hash Table找出同一纵向位置砖缝最多的地方即可。墙的总行数减同一纵向位置的最大砖缝数即是最少需要凿穿的砖块数。具体代码：

class Solution {
public:
    int leastBricks(vector>& wall) {
        unordered_map hash;
        int maxCnt = 0;
        
        for(int i=0; i