问题描述
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:
- 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.
问题分析
问题要求穿过最少的砖块，那么对应的，我们要找到拥有共同前缀长度的最大行数。很自然的，我们使用prefix sums来求解
具体代码如下
示例代码
func leastBricks(wall [][]int) int {
// prefix sum ==> count
m := make(map[int]int)
// maxfreq为最大prefix sum出现的次数
rows, maxfreq := len(wall), 0
for row := 0; row < rows; row++ {
sum := 0
// 计算每一行的prefix sums
for col := 0; col < len(wall[row])-1; col++ {
sum += wall[row][col]
m[sum]++
// 在计算过程中，不断更新maxfreq
if m[sum] > maxfreq {
maxfreq = m[sum]
}
}
}
return rows - maxfreq
}