数组中连续最大和实现

/*auther:hansongjiang*/#include<iostream>#include<vector>using namespace std;//http://www.cnblogs.com/CCBB/archive/2009/04/25/1443455.html//四种方法，一个一个写//第一个O(N*N)  */从第i开始累加的最大的值int getMax0(int a[],int len){int maxSum=0;for(int i=0;i<len;i++){int thisSum=0;for(int j=i;j<len;j++){    thisSum+=a[j];if(thisSum>maxSum){maxSum=thisSum;}}}  return maxSum;}//递归算法思路，就是分而治之，如何分，最简单的二分思路，下回写个二分查找。int getMax1(int a[],int low,int high){if(low==high){if(a[low>0]){return a[low];}else{  return 0;}}int mid=(high+low)/2;//左边low到mid的最大值int left=getMax1(a,low,mid);int right=getMax1(a,mid+1,high);//求左右包括a[mid]的最大值和int  maxSumleft=0;int thisSumleft=0;int j=mid;while(j>=low){    thisSumleft+=a[j];if(maxSumleft<thisSumleft)maxSumleft=thisSumleft;   j--;} j=mid+1; thisSumleft=maxSumleft; while(j<=high){thisSumleft+=a[j];if(thisSumleft>maxSumleft){maxSumleft=thisSumleft;}  j++;}return maxSumleft;}     //o(n)    int getMax2(int a[],int len){     int sum=0; int tempSum=0; for(int i=0;i<len;i++) {         tempSum+=a[i]; if(tempSum<0) tempSum=0; if(tempSum>sum) { sum=tempSum;   }         }    return sum;} void main() { int a[]={3,5,6,-9,1}; int sum=getMax0(a,5); cout<<sum<<endl; sum=getMax2(a,5); cout<<"效率最高的结果"<<sum<<endl; sum=getMax1(a,0,4); cout<<"nlogn"<<sum;       }