数组子数组系列问题的总结

1）最大连续子数组和

     一个有N个元素的整型数组arr，有正有负，数组中连续一个或多个元素组成一个子数组，这个数组当然有很多子数组，求子数组之和的最大值。例如：[0，-2，3，5，-1，2]应返回9，[-9，-2，-3，-5，-3]应返回-2。

     对于O(N³)和O(N²)的解法就不再赘述，这里是分治法（NlogN）和O（N）的实时解法。

1.1分治法

#include <iostream>
#include <vector>
using namespace std;
int MaxSub(vector<int>::const_iterator &,
vector<int>::const_iterator &);
int Max3(int, int,int);
int main(){
    int Length,x,MaxSubsum;
    vector<int> Arr;
    cout<<"Input Length of the array:";
    cin>>Length;
    cout<<"Input the arry:";
    for(int i=1;i<=Length;i++){
       cin>>x;
       Arr.push_back(x);
    }
    vector<int>::const_iterator beg=Arr.begin(),end=Arr.end()-1;
    MaxSubsum=MaxSub(beg,end);
    cout<<"Maximum Summation of subArry is from "
        <<beg-Arr.begin()+1<<
        " to "<<end-Arr.begin()+1<<":"<<endl<<MaxSubsum<<endl;
    system("pause");
    return 0;
}

int MaxSub(vector<int>::const_iterator &beg,
    vector<int>::const_iterator &end){
    int MaxLeft,MaxRight,Max,MaxLeftRight;
    vector<int>::const_iterator mid1,mid2,itr;
    mid1=beg+(end-beg)/2;
    mid2=mid1+1;
    if (beg==end) return *beg;

    MaxLeft=MaxSub(beg,mid1);
    MaxRight=MaxSub(mid2,end);
    for(itr=beg,MaxLeftRight=0;itr!=end+1;itr++)
        MaxLeftRight+=*itr;
    Max=Max3(MaxLeft,MaxRight,MaxLeftRight);
    if(Max==MaxLeft)
    end=mid1;
    else if(Max==MaxRight)
    beg=mid2;

    return Max;
}

int Max3(int a, int b,int c){
    if(a>b) return a>c?a:c;
    else return b>c?b:c;
}

 

1.2 实时算法，时间复杂度复杂度O(N)

#include <iostream>
#include <vector>
int main(){
    int Length,x,MaxSubsum,CurrSubsum=0;
    vector<int> Arr;
    vector<int>::iterator beg,end;
    cout<<"Input Length of the array:";
    cin>>Length;
    cout<<"Input the arry:";
    for(int i=1;i<=Length;i++){
       cin>>x;
       Arr.push_back(x);
    }
    beg=Arr.begin(),end=Arr.end()-1;
    MaxSubsum=Arr[0];
    for(vector<int>::iterator itr=Arr.begin();itr!=Arr.end();itr++){
        CurrSubsum+=*itr;
        if(CurrSubsum<0){CurrSubsum=0;beg=itr+1;}
        else if(CurrSubsum>MaxSubsum){MaxSubsum=CurrSubsum;end=itr;}
    }
    cout<<"Maximum Summation of subArry is from "<<beg-Arr.begin()+1<<
        " to "<<end-Arr.begin()+1<<":"<<endl<<MaxSubsum<<endl;
    system("pause");
       return 0;
}

       以上是两种较优的最大子数组和的算法，最小子数组和同理可求。

 

2）最小正子序列和

思路：(1)先求出数组的前缀数组之和，例如：2,-3,-3,7,5=>前缀数组和为2,-1,-4,3,8（它们的索引0 1 2 3 4）

(2)对前缀数组进行排序得到：-4,-1,2,3,8（即是这样的：(-4,3)  (1,1)  (2,0)  (3,3,)  (8,4) )

(3)则最小值一定是在：-4,1,2,3,8（结果一定在当前项减去前面一项，同时满足索引

大于前一项的索引）

 

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define MAX  1000
class Item{
public:
    int Num;
    int Order;
    Item(): Num(0),Order(0){}
};
bool operator<(const Item &a,const Item &b){
    return a.Order<b.Order;
}
int operator-(const Item& a,const Item& b){
    return a.Num-b.Num;
}
bool comp(const Item &a,const Item& b){
     return a.Num<b.Num;
}

int main(){
    int Length,x,MaxSubsum,CurrSubsum=0;
    vector<int> Arr;
    vector<Item> Sum;
    Item item;
    cout<<"Input Length of the array:";
    cin>>Length;
    cout<<"Input the arry:";
    for(int i=0;i<Length;i++){
       cin>>x;
       Arr.push_back(x);
       item.Num=item.Num+x;
       item.Order=i;
       Sum.push_back(item);
    }
    sort(Sum.begin(),Sum.end(),comp);
    int minpositive=Sum[Length-1].Num>MAX?Sum[Length-1].Num:MAX;
    for(vector<Item>::iterator itr=Sum.begin();itr!=Sum.end()-1;itr++)
        if(*itr<*(itr+1)&&*(itr+1)-*itr>0&&*(itr+1)-*itr<minpositive)
            minpositive=*(itr+1)-*itr;
    cout<<"The minimum Positive Summation is:"<<minpositive<<endl;
    system("pause");
}

    最小负子序列和同理可求.算法复杂度O(N+NlogN)=O(NlogN).

 

3 最大子数组乘积：

     求最大连续子序列乘积与最大连续子序列和问题有所不同，因为其中有正有负还有可能有0。假设数组为a[]，直接利用动归来求解，考虑到可能存在负数的情况，我们用Max[i]来表示以a[i]结尾的最大连续子序列的乘积值，用Min[i]表示以a[i]结尾的最小的连续子序列的乘积值，那么状态转移方程为：

       Max[i]=max{a[i], Max[i-1]*a[i], Min[i-1]*a[i]};

       Min[i]=min{a[i], Max[i-1]*a[i], Min[i-1]*a[i]};

      初始状态为Max[0]=Min[0]=a[0]。算法复杂度位O(N).

代码如下：

 1 double max3(double,double,double);
 2 double min3(double,double,double);
 3 double max3(double a,double b,double c){
 4     if(a>b) return a>c?a:c;
 5     else return b>c?b:c;
 6 }
 7 double min3(double a,double b,double c){
 8     if(a<b) return a<c?a:c;
 9     else return b<c?b:c;
10 }
11 
12 int main()
13 {
14    int n,j;
15    double MaxMupl;
16    double *Arr,*Max,*Min;
17    printf("Input the size of the array:");
18    scanf("%d",&n);
19    Arr=(double *)malloc(sizeof(double)*n);
20    Max=(double *)malloc(sizeof(double)*n);
21    Min=(double *)malloc(sizeof(double)*n);  
22    for ( j=0;j<n;j++)
23       scanf("%lf",Arr+j);
24    Max[0]=Min[0]=MaxMupl=Arr[0];
25    for (int i=1;i<n;i++){
26      Max[i]=max3(Arr[i],Max[i-1]*Arr[i],Min[i-1]*Arr[i]);
27      Min[i]=min3(Arr[i],Max[i-1]*Arr[i],Min[i-1]*Arr[i]);
28      if(Max[i]>MaxMupl) MaxMupl=Max[i];
29    }
30    if(MaxMupl>0){
31        if(MaxMupl==(int)MaxMupl) printf("%d",(int)MaxMupl);
32        else printf("%.2lf",MaxMupl);
33    }
34    else printf("-1\n");
35    system("pause");
36    free(Arr);
37    free(Max);
38    free(Min);
39    return 0;
40 }