1. 程式人生 > >554. Brick Wall 磚墻

554. Brick Wall 磚墻

his out post ges cal cross rom put ros

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:技術分享圖片

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.

  1. class Solution:
  2. def leastBricks(self, wall):
  3. """
  4. :type wall: List[List[int]]
  5. :rtype: int
  6. """
  7. # 穿過最少的磚塊數,即:磚塊行數 - 最大的縫隙數量
  8. # 保存每兩塊磚之間縫隙的Index,找出縫隙數量最多的Index
  9. m = {}
  10. maxGap = 0
  11. for row in wall:
  12. gapIndex = 0
  13. for i in range(0, max(0, len(row) - 1)):
  14. gapIndex += row[i]
  15. num = m.get(gapIndex, 0) + 1
  16. m[gapIndex] = num
  17. maxGap = max(maxGap, num)
  18. return len(wall) - maxGap
  19. wall = [
  20. [1, 2, 2, 1],
  21. [3, 1, 2],
  22. [1, 3, 2],
  23. [2, 4],
  24. [3, 1, 2],
  25. [1, 3, 1, 1]
  26. ]
  27. s = Solution()
  28. res = s.leastBricks(wall)
  29. print(res)





來自為知筆記(Wiz)

554. Brick Wall 磚墻