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  • 最大子数组问题

    1. 问题描述

    对于数组(如下),求解其最大子数组.

       

    结果为:

    2. 算法设计

    采用递归的方法求解

    A. 求解数组左半部分的最大子数组

    B. 求解数组右半部分的最大子数组

    C. 求解整个数组的最大子数组

    D. 比较A,B,C求出的结果,选出一个最大值,即为最终结果.

    3. 数据结构设计

    A. 中间结果的三元组,(子数组下标,子数组上标,子数组和)

    1    class Tuple
    2     {
    3     public:
    4         int low, high, sum;
    5         Tuple(int l = 0, int h = 0, int s = 0):low(l), high(h), sum(s) { }
    6     };

    B. 数组元素

    1  std::vector<int> m_array;

    4. 算法实现

    算法实现文件: calc_max_subarray.h

      1 #include <iostream>
      2 #include <vector>
      3 #include <cassert>
      4 #include <fstream>
      5 
      6 using namespace std;
      7 
      8 namespace nsp_subarray
      9 {
     10     class Tuple
     11     {
     12     public:
     13         int low, high, sum;
     14         Tuple(int l = 0, int h = 0, int s = 0):low(l), high(h), sum(s) { }
     15     };
     16 
     17     class SubArray
     18     {
     19     private:
     20         std::vector<int> m_array;
     21     public: 
     22         SubArray() { m_array.clear(); }
     23 
     24         virtual ~SubArray(){}
     25 
     26         void init_data(string fileName)
     27         {
     28             ifstream in(fileName.c_str());
     29             while(!in.eof())
     30             {
     31                 int e = 0;
     32                 in >> e;
     33                 m_array.push_back(e);
     34             }
     35         }
     36 
     37         inline int get_array_size() { return m_array.size(); }
     38 
     39         Tuple find_max_crossing_subarray(int low, int mid, int high)
     40         {
     41             int m_leftSum = INT_MIN;
     42             int m_rightSum = INT_MIN;
     43 
     44             int m_maxLeft = mid;
     45             int m_maxRight = mid + 1;
     46 
     47             int sum = 0;
     48 
     49             for (int i = mid; i >= low; i--)
     50             {
     51                 sum += m_array[i];
     52                 if (sum > m_leftSum)
     53                 {
     54                     m_leftSum = sum;
     55                     m_maxLeft = i;
     56                 }
     57             }
     58 
     59             sum = 0;
     60             for (int i = mid + 1; i <= high; i++)
     61             {
     62                 sum += m_array[i];
     63                 if (sum > m_rightSum)
     64                 {
     65                     m_rightSum = sum;
     66                     m_maxRight = i;
     67                 }
     68             }
     69 
     70             return Tuple(m_maxLeft, m_maxRight, m_leftSum + m_rightSum);
     71         }
     72 
     73         Tuple find_maximum_subarray(int low, int high)
     74         {
     75             if (low == high)
     76             {
     77                 return Tuple(low, high, m_array[low]);
     78             } 
     79             else if (low < high)
     80             {
     81                 int mid = (low + high) / 2.0;
     82                 Tuple tLeft = find_maximum_subarray(low, mid);
     83 
     84                 Tuple tRight = find_maximum_subarray(mid + 1, high);
     85 
     86                 Tuple tCross = find_max_crossing_subarray(low, mid, high);
     87 
     88                 if (tLeft.sum >= tRight.sum && tLeft.sum >= tCross.sum)
     89                 {
     90                     return tLeft;
     91                 } 
     92                 else if (tRight.sum >= tLeft.sum && tRight.sum >= tCross.sum)
     93                 {
     94                     return tRight;
     95                 }
     96                 else
     97                 {
     98                     return tCross;
     99                 }
    100             }
    101         }
    102     };
    103 };

    main.cpp

     1 #include<iostream>
     2 #include "calc_max_subarray.h"
     3 
     4 using namespace std;
     5 using namespace nsp_subarray;
     6 
     7 int main()
     8 {
     9     SubArray a;
    10     Tuple t;
    11 
    12     a.init_data("A.txt");
    13 
    14     t = a.find_maximum_subarray(0, a.get_array_size() - 1);
    15 
    16     return 0;
    17 }

    6. 文件内容

    文件为: A.txt,内容如下.

    13 -3 -25 20 -3 -16 -23 18 20 -7 12 -5 -22 15 -4 7

    链接:http://pan.baidu.com/s/1gdz35gn 密码:0id2

    https://github.com/LeungGeorge/IntroToAlgo/tree/master/ch04.%E5%88%86%E6%B2%BB%E7%AD%96%E7%95%A5/04.1.calc_max_subarray

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  • 原文地址:https://www.cnblogs.com/BigBigLiang/p/4999251.html
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