LeetCode.554 Brick Wall

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:

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.

分析：

class Solution {
    public int leastBricks(List<List<Integer>> wall) {
        //给定List集合来表示一面墙，每块转的厚度一样，宽度不同，你需要从最上面切一条线，其切过的砖的数量最少。
        //注意：不能从墙的两头切，即不能以最后的PreSum作为判断条件。
        //从两块砖之间切过，表示没有经过砖。砖的行不超过10000，总的砖不超过20000。
        //思路：求每行的PreSum，切过最少的数量一定是每行PreSum相同且数量最多的，总的行数减去PreSum的数量即为结果
        HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
        int maxFre=0;
        for(int i=0;i<wall.size();i++){
            int preSum=0;
            //求preSum,最后一个Sum不作为判断标准
            for(int j=0;j<wall.get(i).size()-1;j++){
                preSum+=wall.get(i).get(j);
                hm.put(preSum,hm.getOrDefault(preSum,0)+1);
                
                //求出现次数最多的
                maxFre=Math.max(maxFre,hm.get(preSum));
            }
        }
        
        return wall.size()-maxFre;
    }
}