（Java）leetcode-554 Brick Wall

topic

【砖墙】
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.

Thinking

对墙中所有可能的缝隙长度（该缝隙到达最左边缘的长度）进行计数。
某长度以及对应出现的次数作为<key,value>维护到map中。
最后出现次数最多的长度就是画线的最佳位置（说明此位置缝隙最多，换言之，穿过的砖头最少）

Code

class Solution {

    public int leastBricks(List<List<Integer>> wall) {
       if(wall.size() == 0) return 0;      
        int count = 0;
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        // 遍历每一行
        for(List<Integer> list : wall){
            int length = 0;
            // 对所有出现的长度计数并更新到map中
            for(int i = 0; i < list.size() - 1; i++){ //不考虑最右边的砖
                length += list.get(i);
                map.put(length, map.getOrDefault(length, 0) + 1);
                // 更新穿过最多的缝隙数
                count = Math.max(count, map.get(length));
            }
        }
        // 穿过的砖数 = 总行数 - 穿过的缝隙数
        return wall.size() - count; 
    }
}

Present the results