zoukankan      html  css  js  c++  java
  • 554. Brick Wall 砖墙

    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. 

    1. class Solution:
    2. def leastBricks(self, wall):
    3. """
    4. :type wall: List[List[int]]
    5. :rtype: int
    6. """
    7. # 穿过最少的砖块数,即:砖块行数 - 最大的缝隙数量
    8. # 保存每两块砖之间缝隙的Index,找出缝隙数量最多的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)







  • 相关阅读:
    MySQL之数据库结构优化
    MySQL之索引
    Spring之单元测试
    Spring之IOC容器加载初始化的方式
    LeetCode之Sort List
    [译]Java 垃圾回收的监控和分析
    [译]Java垃圾回收器的类型
    [译]Java垃圾回收是如何工作的
    [译]Java 垃圾回收介绍
    JSP之项目路径问题(${pageContext.request.contextPath},<%=request.getContextPath()%>以及绝对路径获取)
  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/8419856.html
Copyright © 2011-2022 走看看