leetcode554. Brick Wall

题目要求

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.

现在有一个二维数组来表示一面墙，其中二维数组中每一个值代表某一行的一颗砖块的长度。现在会画一条垂直的线，问最少需要跨过多块砖。

思路和代码

本质上就是记录一下经过的所有砖块缝隙，然后遇到重复的砖块缝隙就将该为止的砖块缝隙数量加一。最后穿过越多的砖块缝隙代表所需要穿过的砖块数量越少。

public int leastBricks(List<List<Integer>> wall) {  
    Map<Integer, Integer> map = new HashMap<>();  
    int row = wall.size();  
    int maxEdge = 0;  
    for (List<Integer> bricks : wall) {  
        int sumOfBrick = 0;  
        for (int i = 0; i < bricks.size() - 1; i++) {  
            int brick = bricks.get(i);  
            sumOfBrick += brick;  
            int edge = map.getOrDefault(sumOfBrick, 0) + 1;  
            map.put(sumOfBrick, edge);  
            maxEdge = Math.max(edge, maxEdge);  
        }  
    }  
    return row - maxEdge;  
}