Leetcode-554. Brick Wall

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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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.

这个题目有一定的tick，当我们遍历每一行的时候可以用一个hashmap保存：当前累计宽度为多少的值，出现了多少次。那么跨越的数量就是层数-出现的次数，求一个最小值就可以了。

public class Solution {
    public int leastBricks(List> wall) {
        Map widths = new HashMap();
        int minNum = wall.size();
        for(List rows : wall){
            int sum = 0;
            for(int i = 0; i < rows.size() - 1; i ++){
                sum += rows.get(i);
                int num = widths.getOrDefault(sum,0);
                widths.put(sum,++num);
                minNum = Math.min(minNum, wall.size() - num);
            }
        }
        
        return minNum;
    }
}