算法设计与分析--求最大子段和问题（蛮力法分治法动态规划法 C++实现

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算法设计与分析--求最大子段和问题

问题描述：

**给定由*n*个整数组成的序列(*a*1,*a*2, …,*an*)，求该序列形如**

![20130513112721129][]

的子段和的最大值，当所有整数均为负整数时，其最大子段和为0。

利用蛮力法求解：

int maxSum(int a[],int n){ int maxSum = 0; int sum = 0; for(int i = 0; i < n; i++) //从第一个数开始算起 {  for(int j = i + 1; j < n; j++)//从i的第二个数开始算起  {   sum = a[i];   a[i]  += a[j];   if(a[i] > sum)   {    sum = a[i];  //每一趟的最大值   }  }  if(sum > maxSum)  {   maxSum = sum;  } } return maxSum;}

利用分治法求解：

int maxSum(int a[],int left, int right){ int sum = 0; if(left == right) //如果序列长度为1，直接求解 {  if(a[left] > 0) sum = a[left];  else sum = 0; } else  {  int center = (left + right) / 2; //划分  int leftsum = maxSum(a,left,center); //对应情况1，递归求解  int rightsum = maxSum(a, center + 1, right);//对应情况2， 递归求解  int s1 = 0;  int lefts = 0;  for(int i = center; i >= left; i--) //求解s1  {   lefts += a[i];   if(lefts > s1) s1 = lefts; //左边最大值放在s1  }  int s2 = 0;   int rights = 0;  for(int j = center + 1; j <= right; j++)//求解s2  {   rights += a[j];   if(rights > s2) s2 =rights;  }  sum = s1 + s2;    //计算第3钟情况的最大子段和  if(sum < leftsum) sum = leftsum; //合并，在sum、leftsum、rightsum中取最大值  if(sum < rightsum) sum = rightsum; } return sum;}

**利用动态规划法求解：**

int DY_Sum(int a[],int n){ int sum = 0; int *b = (int *) malloc(n * sizeof(int)); //动态为数组分配空间 b[0] = a[0]; for(int i = 1; i < n; i++) {  if(b[i-1] > 0)   b[i] = b[i - 1] + a[i];  else   b[i] = a[i]; } for(int j = 0; j < n; j++) {  if(b[j] > sum)   sum = b[j]; } delete []b;  //释放内存 return sum;}

完整测试程序：

#include<iostream>#include<time.h>#include<Windows.h>using namespace std;#define MAX 10000int BF_Sum(int a[],int n)   { int max=0;      int sum=0;         int i,j; for (i=0;i<n-1;i++)         {           sum=a[i];            for(j=i+1;j<n;j++)              {          if(sum>=max)                   {                                             max=sum;                   }     sum+=a[j];           }     }     return max;}    int maxSum1(int a[],int left, int right){ int sum = 0; if(left == right) //如果序列长度为1，直接求解 {  if(a[left] > 0) sum = a[left];  else sum = 0; } else  {  int center = (left + right) / 2; //划分  int leftsum = maxSum1(a,left,center); //对应情况1，递归求解  int rightsum = maxSum1(a, center + 1, right);//对应情况2， 递归求解  int s1 = 0;  int lefts = 0;  for(int i = center; i >= left; i--) //求解s1  {   lefts += a[i];   if(lefts > s1) s1 = lefts; //左边最大值放在s1  }  int s2 = 0;   int rights = 0;  for(int j = center + 1; j <= right; j++)//求解s2  {   rights += a[j];   if(rights > s2) s2 =rights;  }  sum = s1 + s2;    //计算第3钟情况的最大子段和  if(sum < leftsum) sum = leftsum; //合并，在sum、leftsum、rightsum中取最大值  if(sum < rightsum) sum = rightsum; } return sum;}int DY_Sum(int a[],int n){ int sum = 0; int *b = (int *) malloc(n * sizeof(int)); //动态为数组分配空间 b[0] = a[0]; for(int i = 1; i < n; i++) {  if(b[i-1] > 0)   b[i] = b[i - 1] + a[i];  else   b[i] = a[i]; } for(int j = 0; j < n; j++) {  if(b[j] > sum)   sum = b[j]; } delete []b;  //释放内存 return sum;}int main(){ int num[MAX]; int i; const int n = 40; LARGE_INTEGER begin,end,frequency; QueryPerformanceFrequency(&frequency); //生成随机序列 cout<<"生成随机序列："; srand(time(0)); for(int i = 0; i < n; i++) {  if(rand() % 2 == 0)   num[i] = rand();  else   num[i] = (-1) * rand();  if(n < 100)   cout<<num[i]<<" "; } cout<<endl; //蛮力法// cout<<"\n蛮力法:"<<endl; cout<"最大字段和："; QueryPerformanceCounter(&begin); cout<<BF_Sum(num,n)<<endl; QueryPerformanceCounter(&end); cout<<"时间:"  <<(double)(end.QuadPart - begin.QuadPart) / frequency.QuadPart  <<"s"<<endl; cout<<"\n分治法:"<<endl; cout<"最大字段和："; QueryPerformanceCounter(&begin); cout<<maxSum1(num,0,n)<<endl; QueryPerformanceCounter(&end); cout<<"时间:"  <<(double)(end.QuadPart - begin.QuadPart) / frequency.QuadPart  <<"s"<<endl; cout<<"\n动态规划法:"<<endl; cout<"最大字段和："; QueryPerformanceCounter(&begin); cout<<DY_Sum(num,n)<<endl; QueryPerformanceCounter(&end); cout<<"时间:"  <<(double)(end.QuadPart - begin.QuadPart) / frequency.QuadPart  <<"s"<<endl; system("pause"); return 0;}

测试结果：

![20130513214725440][]

[https_blog.csdn.net_jiangjunshow]: https://blog.csdn.net/jiangjunshow/article/details/77338485
[20130513112721129]: /images/20220317/ec5c8b38cc784084a32625aef9fe58e2.png
[20130513214725440]: /images/20220317/c565321c89b147eda6bac1c7799beb61.png