DivideAndConquer-最大子数组问题

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串行策略

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Maximum_subarray
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] originArray = { 13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7};
            MaxSub maxsub = new MaxSub(originArray);
            var res = maxsub.Result;
            Console.Write("The Maximum subarray is : ");
            for (int i = res.Item1; i <= res.Item2; i++)
            {
                Console.Write("{0} ",originArray[i]);
            }
            Console.Write("\nAnd the sum is : {0}",res.Item3);
            Console.Write("\n");
            Console.Read();         
        }
    }
    class MaxSub
    {
        public MaxSub(int [] originarray)
        {
            OringinArray = originarray;
        }
        public int [] OringinArray
        {
            get;
            private set;
        }
        public Tuple<int,int,int> Result
        {
            get
            { return Generate(); }
            private set{}
        }
        private Tuple<int,int,int> Generate()
        {
            int firstIdx, lastIdx, sum;
            firstIdx = lastIdx = sum = 0;
            var turple = Tuple.Create(firstIdx, lastIdx, sum);
            for (int i = 0; i < OringinArray.Length;i++ )
            {
                sum += OringinArray[i];
                lastIdx = i;
                if (sum < OringinArray[i])
                {
                    firstIdx = i;
                    sum = OringinArray[i];
                }
                if (sum > turple.Item3)
                {
                    turple = Tuple.Create(firstIdx,lastIdx,sum);
                }
                //Console.Write("{0} {1} {2}\n", firstIdx, lastIdx, sum);
            }
            return turple;
        }

    }
}

分治策略

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace MaxSubArray_DivAndCon
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] originArray = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
            MaxSubArrayDAC maxSub = new MaxSubArrayDAC(originArray);
            var res = maxSub.Result;
            Console.Write("The Maximum subarray is : ");
            for (int i = res.Item1; i <= res.Item2; i++)
            {
                Console.Write("{0} ", originArray[i]);
            }
            Console.Write("\nAnd the sum is : {0}", res.Item3);
            Console.Write("\n");
            Console.Read();
        }
    }
    class MaxSubArrayDAC
    {
        public MaxSubArrayDAC(int [] originArray)
        {
            OriginArray = originArray;
        }
        public int[] OriginArray
        {
            get;
            private set;
        }
        public Tuple<int, int, int> Result
        {
            get{return Generate(0,OriginArray.Length-1);}
            private set { }
        }
        private Tuple<int,int,int> Generate(int leftIdx,int rightIdx)
        {
            if (leftIdx == rightIdx)
                return Tuple.Create(leftIdx, rightIdx, OriginArray[leftIdx]);
            else
            {
                int midIdx = (leftIdx + rightIdx) / 2;
                var leftTurple = Generate(leftIdx,midIdx);
                var rightTurple = Generate(midIdx+1,rightIdx);
                var midTurple = MaxCrossingSubArray(leftIdx, midIdx, rightIdx);
                if (leftTurple.Item3 > midTurple.Item3 && leftTurple.Item3 > rightTurple.Item3)
                    return leftTurple;
                else if (midTurple.Item3 > leftTurple.Item3 && midTurple.Item3 > rightTurple.Item3)
                    return midTurple;
                else
                    return rightTurple;
            }

        }
        private Tuple<int,int,int> MaxCrossingSubArray(int firIdx, int midIdx, int lastIdx)
        {
            int leftSum, rightSum,temp;
            leftSum = rightSum = temp = 0;
            int leftIdx, rightIdx;
            leftIdx = rightIdx = midIdx;
            for (int i = midIdx; i >= firIdx; i--)
            {
                temp += OriginArray[i];
                if (temp > leftSum)
                {
                    leftSum = temp;
                    leftIdx = i;
                }
            }
            temp = 0;
            for (int i = midIdx+1; i <= lastIdx; i++)
            {
                temp += OriginArray[i];
                if (temp > rightSum)
                {
                    rightSum = temp;
                    rightIdx = i;
                }
            }
            return Tuple.Create(leftIdx, rightIdx, leftSum + rightSum);
        }
    }
}


IgH EtherCAT 从入门到精通:第 29 章 实战:EoE 远程维护通道搭建 摘要:本章详细介绍了基于EoE(Ethernet over EtherCAT)协议的远程维护通道搭建方案。主要内容包括:1)网络架构设计,采用管理网络与EoE子网隔离的方案;2)从站EoE支持确认与IP地址配置,提供API和命令行两种方式;3)路由/交换两种部署模式,重点推荐路由模式并给出自动化配置脚本;4)SSH远程访问配置,包括直接连接和跳板访问两种方式;5)安全加固措施,如防火墙规则、SSH密钥认证等;6)故障恢复机制与性能注意事项。该方案实现了通过EtherCAT网络对从站设备的远程SSH访问、文件 阅读详情

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学号:16340008 题目:312. Burst Balloons Question: Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons....

c980129的博客 344

leetcode 53.最大自子序和

Dynamic Programming There is a nice introduction to the DP algorithm in this Wikipedia article. The idea is to maintain a running maximum smax and a current summation sum. When we visit each num in nu...

aikudexue的博客 252

算法实验报告 排序类

E1-1 Problem 在数据库中查找对应key的index V1 思路: 直接使用遍历的方法,每次查找都遍历一遍数组,如果遍历直至结束都没有找到与k相同的数字,输出-1 时间复杂度 时间复杂度:每次遍历寻找的时间复杂度是O(n) 细节&问题 第一次用了很多库函数来处理输入gets()、atoi()、double log10(double) 等等,但是出现了很多的runtime err...

Mrs_Two的博客 495

面试算法53. Maximum Subarray

Leetcode 53. Maximum Subarray 解法1. 使用dp, O(n) 解法2. 使用divide and conquer 对于每一个子数组都分成L,R两部分,分别寻找L,R中最大的子序列 a. L中的最大 b. R中的最大 c. 横穿L,R的最大(与平面中的点同理) T(n) = 2T(n/2) + n T(n) = O(nlogn) class Solution { ...

博客 220

Leecode654 最大二叉树

【代码】Leecode654 最大二叉树。

Stanly_的博客 169

Maximum Subarray

原题请戳Maximum Subarray 大意为:求给定数组的最大子数组的元素之和。 本题可用两种方法解题,分别为:动态规划和divide and conquer,本文先介绍动态规划解法,divide and conquer解法见下篇博客 本文算是一道比较经典的动态规划题目了,与暴力求解(穷举法)相比,可将时间复杂度降为O(n),空间复杂度仅为O(1) 采用“局部最优和全局最优解法” 顾

chouchou680的博客 285

leetCode No.53 Maximum Subarray

题目 Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has t

Andy_Shan的博客 803

[勇者闯LeetCode] 53. Maximum Subarray

[勇者闯LeetCode] 53. Maximum SubarrayDescription Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,

fanguzhai的博客 393

53. Maximum Subarray

最大和的子数组,leetcode, 动态规划,分治法,divide and conquer

New coder, Step further 221

[LeetCode] Maximum Subarray Sum

Dynamic Programming There is a nice introduction to the DP algorithm in thisWikipedia article. The idea is to maintain a running maximumsmaxand a current summationsum. When ...

weixin_34235105的博客 96
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