zoukankan      html  css  js  c++  java
  • Brick Wall

    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.

    Analyses: look over the entire list, record the length of each brick, find the most frequent brick length, then return the list length  - that frequency. 

     1 public class Solution {
     2     public int leastBricks(List<List<Integer>> wall) {
     3         int n = wall.size();
     4         if (n == 0) return 0;
     5         
     6         int mostCount = 0;
     7         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
     8         for (List<Integer> row : wall) {
     9             int rowSize = row.size();
    10             int currentLength = 0;
    11             // ignore the last column
    12             for (int i = 0; i < rowSize - 1; i++) {
    13                 currentLength += row.get(i);
    14                 map.put(currentLength, map.getOrDefault(currentLength, 0) + 1);
    15                 mostCount = Math.max(mostCount, map.get(currentLength));
    16             }
    17         }
    18         return n - mostCount;
    19     }
    20 }
  • 相关阅读:
    oracle数据段详解
    监听静态注册与动态注册
    Oracle网络相关概念与常用配置文件
    pycharm社区版安装及遇到的问题
    强化学习-K摇臂赌博机
    概率图模型
    半监督学习
    卷积神经网络
    递归神经网络
    玻尔兹曼机及其相关模型
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6752369.html
Copyright © 2011-2022 走看看