求数组的最大子段和(O(N^3)-->O(N^2)-->O(NlogN)-->O(N))

最新推荐文章于 2023-10-21 23:43:37 发布

原创最新推荐文章于 2023-10-21 23:43:37 发布 · 1.6k 阅读

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#methods #algorithm #include #2010 #date

本文介绍了一种寻找数组中最大子数组和的有效算法，从原始的三次嵌套循环逐步优化到二次、一次嵌套循环及分治递归，最终采用动态规划实现线性时间复杂度的高效解决方案。

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三层嵌套循环-->二层嵌套循环-->分治递归-->动态规划

/******************************************************************* * * DESCRIPTION: Algorithm 20100117 MaxSubArraySum() * * AUTHOR:Neesky----O(N^3)-->O(N^2)-->O(NlogN)-->O(N) * * DATE:2010-1-17 * *******************************************************************/ /** include files **/ #include <iostream> using namespace std; /// <summary> /// The Ordinary Method to get the max sub arrays /// </summary> /// <param name="A"></param> /// <param name="N"></param> /// <returns></returns> int MaxSubArraySumON3_inArray(int A[],int N) { int Sum=0,MaxSum=0; for(int i=0;i<N;++i) { for(int j=i;j<N;++j) { Sum=0; for(int k=i;k<=j;++k)/*O(N^3)*/ Sum+=A[k]; MaxSum=Sum>MaxSum?Sum:MaxSum; } } return MaxSum; } /*Optimize the O(N^3) to O(N^2)*/ int MaxSubArraySumON2_inArray(int A[],int N) { int Sum=0,MaxSum=0; for(int i=0;i<N;++i) { Sum=0;/*Just Optimize it*/ for(int j=i;j<N;j++) { Sum+=A[j]; MaxSum=Sum>MaxSum?Sum:MaxSum; } } return MaxSum; } /// <summary> /// Divide and Conquer(Get the Max Sub Array Sum) :O(NlogN)[分治] /// </summary> /// <param name="A"></param> /// <param name="N"></param> /// <param name="i"></param> /// <param name="j"></param> /// <returns></returns> int MaxSubArraySumDC_inArray(int A[],int left,int right) { if(left==right)return A[left]>0?A[left]:0; /*Exit Of Recursive*/ int center=(left+right)/2; int leftSum=MaxSubArraySumDC_inArray(A,left,center); int rightSum=MaxSubArraySumDC_inArray(A,center+1,right); int leftSecSum=0,leftSecMaxSum=0; /*left section*/ for(int i=center;i>=left;--i) { leftSecSum+=A[i]; leftSecMaxSum=leftSecSum>leftSecMaxSum?leftSecSum:leftSecMaxSum; } int rightSecSum=0,rightSecMaxSum=0; /*right section*/ for(int i=center+1;i<=right;++i) { rightSecSum+=A[i]; rightSecMaxSum=rightSecSum>rightSecMaxSum?rightSecSum:rightSecMaxSum; } int CenterSum=leftSecMaxSum+rightSecMaxSum; /*get the maxSum from 3 sums*/ int resSum=CenterSum; if(resSum<leftSum)resSum=leftSum; if(resSum<rightSum)resSum=rightSum; return resSum; } int MaxSubArraySumDC_inArray(int A[],int N) { return MaxSubArraySumDC_inArray(A,0,N-1); } /// <summary> /// Dynamic Plan(Get the Max Sub Array Sum) :O(N)[动态规划] /// </summary> /// <param name="A"></param> /// <param name="N"></param> /// <returns></returns> int MaxSubArraySumDP_inArray(int A[],int N) { /*b is Max(1<=i<=j)(Sum(1<=i<=n)A[i])*/ int sum=0,b=0; /*Just 2 variable for DP*/ for(int i=0;i<N;++i) { if(b>0)b=b+A[i];/*Update the b variable*/ else b=A[i]; if(b>sum)sum=b; /*Update the sum variable*/ } return sum; } /// <summary> /// The Main Programming /// </summary> /// <param name="argc"></param> /// <param name="argv"></param> /// <returns></returns> int main (int argc, char *argv[]) { int A[6]={-2,11,-4,13,-5,-2}; cout<<"/nF3 Methods( O(N^3) )-->"<<MaxSubArraySumON3_inArray(A,6)<<endl; cout<<"/nF2 Methods( O(N^2) )-->"<<MaxSubArraySumON2_inArray(A,6)<<endl; cout<<"/nDP Methods( O(N) )-->"<<MaxSubArraySumDP_inArray(A,6)<<endl; cout<<"/nDC Methods(O(NlogN))-->"<<MaxSubArraySumDC_inArray(A,6)<<endl; return(0); }

mingw32-make: Nothing to be done for `all'.
Debug/Algorithm20100117.exe 

F3 Methods( O(N^3) )-->20

F2 Methods( O(N^2) )-->20

DP Methods( O(N)   )-->20

DC Methods(O(NlogN))-->20