Path sum: two ways

In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down, is indicated in bold red and is equal to 2427.

         
131 673 234 103 18
201 96 342 965 150
630 803 746 422 111
537 699 497 121 956
805 732 524 37 331

Find the minimal path sum, in matrix.txt (right click and “Save Link/Target As…”), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by only moving right and down.


路径和:两个方向

在如下的5乘5矩阵中,从左上方到右下方始终只向右或向下移动的最小路径和为2427,由标注红色的路径给出。

         
131 673 234 103 18
201 96 342 965 150
630 803 746 422 111
537 699 497 121 956
805 732 524 37 331

在这个31K的文本文件matrix.txt(右击并选择“目标另存为……”)中包含了一个80乘80的矩阵,求出从该矩阵的左上方到右下方始终只向右和向下移动的最小路径和。

解题

这个题目很简单的

对第0列和第0行的数直接向下加

第0列:data[i][0] = data[i][0] + data[i-1][0]  for i in 1:row - 1

第0行: data[0][i] = data[0][i] + data[0][i-1] for i in 1:col-1

其他情况

for i in 1:row -1

for j in 1:col-1

data[i][j] = data[i][j] + min(data[i-1][j],data[i][j-1])

最后元素data[row-1][col-1]就是最小路径的值。

Python

import time ;
import numpy as np def run():
filename = 'E:/java/projecteuler/src/Level3/p081_matrix.txt'
data = readData(filename)
Path_Sum(data) def Path_Sum(data):
row,col = np.shape(data)
for i in range(1,row):
data[0][i] = data[0][i]+data[0][i-1]
data[i][0] = data[i][0] + data[i-1][0]
for i in range(1,row):
for j in range(1,col):
data[i][j] += min(data[i-1][j],data[i][j-1])
print data[row-1][col-1] def readData(filename):
fl = open(filename)
data =[]
for row in fl:
row = row.split(',')
line = [int(i) for i in row]
data.append(line)
return data
if __name__=='__main__':
t0 = time.time()
run()
t1 = time.time()
print "running time=",(t1-t0),"s" #
# running time= 0.00799989700317 s

参考博客中的读取文件,这个读取文件的思想很好的,自己对于读取文件还不是很熟悉

上个Python程序是按照左上到右下走的

下面java的是按照右下向左上走的

package Level3;

import java.awt.List;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList; public class PE081{ static int[][] grid;
static void run() throws IOException{
String filename = "src/Level3/p081_matrix.txt";
String lineString = "";
ArrayList<String> listData = new ArrayList<String>();
BufferedReader data = new BufferedReader(new FileReader(filename));
while((lineString = data.readLine())!= null){
listData.add(lineString);
}
// 分配大小空间的 定义的grid 没有定义大小
assignArray(listData.size());
// 按照行添加到数组grid中
for(int index = 0,row_counter=0;index <=listData.size() - 1;++index,row_counter++){
populateArray(listData.get(index),row_counter);
}
System.out.println(Path_min(grid)); }
public static int Path_min(int[][] data){
int size = data.length;
for(int i=size -2;i>=0;--i){
data[i][size-1] += data[i+1][size-1];
data[size-1][i] += data[size-1][i+1];
}
for( int index = size -2;index >=0;index--){
for(int innerIndex = size -2;innerIndex >=0;innerIndex--){
data[index][innerIndex] += Math.min(data[index+1][innerIndex],
data[index][innerIndex+1]);
}
}
return data[0][0];
}
// 每行的数据添加到数组中
public static void populateArray(String str,int row){
int counter = 0;
String[] data = str.split(",");
for(int index = 0;index<=data.length -1;++index){
grid[row][counter++] = Integer.parseInt(data[index]);
}
}
public static void assignArray(int no_of_row){
grid = new int[no_of_row][no_of_row];
} public static void main(String[] args) throws IOException{
long t0 = System.currentTimeMillis();
run();
long t1 = System.currentTimeMillis();
long t = t1 - t0;
System.out.println("running time="+t/1000+"s"+t%1000+"ms");
// 427337
// running time=0s38ms
}
}

Project Euler 81:Path sum: two ways 路径和:两个方向的更多相关文章

  1. Project Euler 83:Path sum: four ways 路径和:4个方向

    Path sum: four ways NOTE: This problem is a significantly more challenging version of Problem 81. In ...

  2. Project Euler 82:Path sum: three ways 路径和:3个方向

    Path sum: three ways NOTE: This problem is a more challenging version of Problem 81. The minimal pat ...

  3. Leetcode 931. Minimum falling path sum 最小下降路径和(动态规划)

    Leetcode 931. Minimum falling path sum 最小下降路径和(动态规划) 题目描述 已知一个正方形二维数组A,我们想找到一条最小下降路径的和 所谓下降路径是指,从一行到 ...

  4. 【LeetCode-面试算法经典-Java实现】【064-Minimum Path Sum(最小路径和)】

    [064-Minimum Path Sum(最小路径和)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a m x n grid filled with ...

  5. [LeetCode] Path Sum II 二叉树路径之和之二

    Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given su ...

  6. [LeetCode] Path Sum 二叉树的路径和

    Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all ...

  7. [LeetCode] Binary Tree Maximum Path Sum(最大路径和)

    Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. ...

  8. [LeetCode] 113. Path Sum II 二叉树路径之和之二

    Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given su ...

  9. [LeetCode] 112. Path Sum 二叉树的路径和

    Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all ...

随机推荐

  1. 30个惊人的插件来扩展 Twitter Bootstrap

    Bootstrap Maxlength It is a lightweight plugin that allows detecting the HTML maxlength property of ...

  2. SharedPreferences的基本用法

    获取SharedPreferences的两种方式: 1 调用Context对象的getSharedPreferences()方法 2 调用Activity对象的getPreferences()方法 两 ...

  3. 读取iis日志到sql server

    using Fasterflect; using System; using System.Collections.Generic; using System.Data.SqlClient; usin ...

  4. phpStudy 2016 更新下载,新版支持php7.0

    目标:让天下没有难配的php环境. phpStudy Linux版&Win版同步上线 支持Apache/Nginx/Tengine/Lighttpd/IIS7/8/6 『软件简介』该程序包集成 ...

  5. MSSQL优化之——查看语句执行情况

    MSSQL优化之——查看语句执行情况 在写SQL语句时,必须知道语句的执行情况才能对此作出优化.了解SQL语句的执行情况是每个写程序的人必不可少缺的能力.下面是对查询语句执行情况的方法介绍. 一.设置 ...

  6. Android砖机救活(索爱MT15i)

    前言 接触Android时间长了就想编译一套属于自己的系统,摘取不必要的那些组件,然后刷到手机上,俗话说的好,“常在河 边走,哪有不湿鞋”.果不其然,刷完自己编译的系统手机变砖了,具体情况为 开不开机 ...

  7. centos系统下安装使用composer教程

    Composer 是 PHP 的一个依赖管理工具.它允许你申明项目所依赖的代码库,它会在你的项目中为你安装他们.Composer 不是一个包管理器.是的,它涉及 "packages" ...

  8. oracle11g RAC添加节点

    OS: [root@rac ~]# more /etc/oracle-releaseOracle Linux Server release 5.7 DB: SQL> SELECT * FROM ...

  9. 安装mysql 5.5.14 报错

    提示cmake nod foundyum install cmake 原因是曾经服务器安装过mysql数据库Installing MySQL system tables...101223 14:28: ...

  10. 【Jetlang】一个高性能的Java线程库

    actor  Jetlang 提供了一个高性能的Java线程库,该库是 JDK 1.5 中的 java.util.concurrent 包的补充,可用于基于并发消息机制的应用. .net的MS CCR ...