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

 

posted @ 2015-11-26 22:30  leungcnblogs  阅读(322)  评论(0)  编辑  收藏  举报