最大子段和

问题描述

• 求一个序列的最大子段和即最大连续子序列之和。例如序列[4, -3, 5, -2, -1, 2, 6, -2]的最大子段和为11=[4+(-3)+5+(-2)+(-1)+(2)+(6)]。

 

方法一：分治

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int MaxSubSum(const int arr[],int l,int r){
    int sum=0;
    if(l==r) 
        return arr[l];
    else{
        int mid=l+(r-l)/2;
        int lsum=MaxSubSum(arr,l,mid);
        int rsum=MaxSubSum(arr,mid+1,r);
        int s1=0;
        int lefts=0;
        for(int i=mid; i>=l; i--){
            lefts+=arr[i];
//            s1=max(lefts,s1);
            if(lefts>s1) s1=lefts;
        }
        int s2=0;
        int rights=0;
        for( int i=mid+1; i<=r; i++ ){
            rights+=arr[i];
//            s2=max(rights,s2);
            if(rights>s2) s2=rights;
        }
        sum=s1+s2;
        sum=max(sum,max(lsum,rsum));
        return sum;
    }
}

int main(){
    const int a[] = { 4, -3, 5, -2, -1, 2, 6, -2 };
    int maxSubSum=MaxSubSum(a,0,7);
    cout<<maxSubSum<<endl;


    return  0;
}

方法二：动态规划

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 20;

int main(){
    const int a[] = { 0,4, -3, 5, -2, -1, 2, 6, -2 };
    int b[maxn];
    memset(b,0,sizeof(b));
    int sum=0;
    int len =sizeof(a)/sizeof(int);
    for( int i=1; i<len; i++ ){
        if(b[i-1]>0) b[i]=b[i-1]+a[i];
        else b[i]=a[i];
        sum=max(sum,b[i]);
    }
    cout<<sum<<endl;

    return 0;
}

。。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn = 20;
int sum[maxn];

int main(){
    memset(sum,0,sizeof(sum));
    const int a[] = { 0,4, -3, 5, -2, -1, 2, 6, -2 };
    int len =sizeof(a)/sizeof(int);
    for( int i=1; i<=len; i++ ){
        sum[i]=sum[i-1]+a[i];
    }
    for( int i=0; i<len; i++ ){
        cout<<sum[i]<<" ";
    }
    cout<<endl;
    int ove_min=INF;
    int ove_max=-INF;
    for( int i=0; i<len; i++ ){
        ove_min=min(ove_min,sum[i]);
        ove_max=max(ove_max,sum[i]);
    }
    cout<<ove_max<<" "<<ove_min<<endl;

    return 0;
}