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. silverlight 用户浏览器未安装SL插件问题

    1.在Silverlight启动页面 <%@ Page Language="C#" AutoEventWireup="true" %> <!D ...

  2. C# WinForm设置TreeView选中节点

    这里假定只有两级节点,多级方法类似.遍历节点,根据选中节点文本找到要选中的节点.treeView.SelectedNode = selectNode; /// <summary> /// ...

  3. openerp学习笔记 计算字段支持搜索

    示例1: # -*- encoding: utf-8 -*-import poolerimport loggingimport netsvcimport toolslogger = netsvc.Lo ...

  4. Ztack学习笔记(3)-系统启动分析

    一 系统启动 //OSAL.cvoid osal_start_system( void ) { #if !defined ( ZBIT ) && !defined ( UBIT ) f ...

  5. 1093. Count PAT's (25)

    The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and ...

  6. PAT IO-02 整数四则运算

    /* *PAT IO-02 整数四则运算 *2015-07-30 *作者:flx413 */ #include<stdio.h> int main() { int a, b; scanf( ...

  7. Python支持中文注释

    三处设置,使Python的Eclipse开发环境(使用PyDev)支持中文 - (a)Eclipse的Window菜单Editors设置: Eclipse工具条 -> Window -> ...

  8. iOS开发中常用第三方库的使用和配置-GDataXML

    这篇文章旨在给自己以后需要时能及时的查到,省得每次都去baidu. 1. xml解析库-GDataXML 参考文章:http://blog.csdn.net/tangren03/article/det ...

  9. Java并发编程:Lock(上)

    在上一篇文章中我们讲到了如何使用关键字synchronized来实现同步访问.本文我们继续来探讨这个问题,从Java 5之后,在java.util.concurrent.locks包下提供了另外一种方 ...

  10. 音频播放、录音、视频播放、拍照、视频录制-b

    随着移动互联网的发展,如今的手机早已不是打电话.发短信那么简单了,播放音乐.视频.录音.拍照等都是很常用的功能.在iOS中对于多媒体的支持是非常强大的,无论是音视频播放.录制,还是对麦克风.摄像头的操 ...