所謂最大子序列和問題：
給定整數A1，A2，... ，An(可能有負數)，求∑Ak的最大值（k從i到j），如果所有整數均爲負數，則最大子序列和爲0。
下面通過4個算法來求解最大子序列和問題，每個算法的效率都不一樣且效率越來越好。
算法代碼來自《數據結構與算法分析——C語言描述》第2版
/******算法 1，時間複雜度O(n3) ******/
int MaxSubSequenceSum(const int A[ ],int N)
{
int ThisSum, MaxSum, i, j, k;
MaxSum = 0;
for(i = 0;i < N;i++)
for(j = i;j < N;j++)
{
ThisSum = 0;
for(k = i;k < j;k++)
ThisSum += A[k];
if(ThisSum > MaxSum)
MaxSum = ThisSum;
}
return MaxSum;
}
/******算法 2，時間複雜度O(n2) ******/
int MaxSubSequenceSum(const int A[ ],int N)
{
int ThisSum, MaxSum, i, j;
MaxSum = 0;
for(i = 0;i < N;i++)
{
ThisSum = 0;
for(j = i;j < N;j++)
{
ThisSum += A[j];
if(ThisSum > MaxSum)
MaxSum = ThisSum;
}
}
return MaxSum;
}
/******算法 3,採用分治算法，時間複雜度O(nlogn) ******/
int Max3(int x,int y,int z)
{
return (x>y)?(x>z?x:z):(y>z?y:z);
}
static int MaxSubSum(const int A[ ],int Left,int Right)
//static所修飾的函數，表示該函數只能在本文件(模塊) //中使用，其他文件不能調用該函數
{
int MaxLeftSum, MaxRightSum;
int MaxLeftBorderSum, MaxRightBorderSum;
int LeftBorderSum, RightBorderSum;
int Center, i;
if(Left == Right) /*Base case*/
if(A[Left] > 0)
return A[Left];
else
return 0;
Center = (Left + Right)/2;
MaxLeftSum = MaxSubSum(A,Left,Center);
MaxRightSum = MaxSubSum(A,Center+1,Right);
MaxLeftBorderSum = 0; LeftBorderSum = 0;
for(i = Center;i >= Left;i--)
{
LeftBorderSum += A[i];
if(LeftBorderSum > MaxLeftBorderSum)
MaxLeftBorderSum = LeftBorderSum;
}
MaxRightBorderSum = 0; RightBorderSum = 0;
for(i = Center + 1;i <= Right;i++)
{
RightBorderSum += A[i];
if(RightBorderSum > MaxRightBorderSum)
MaxRightBorderSum = RightBorderSum;
}
return Max3(MaxLeftSum,MaxRightSum,MaxLeftBorderSum + MaxRightBorderSum);
}
int MaxSubSequenceSum(const int A[ ],int N)
{
return MaxSubSum(A,0,N-1);
}
/******算法4 ,時間複雜度O(n) ******/
int MaxSubSequenceSum(const int A[],int N)
{
int ThisSum, MaxSum, i;
ThisSum = MaxSum = 0;
for(i = 0;i < N;i++)
{
ThisSum += A[i];
if(ThisSum > MaxSum)
MaxSum = ThisSum;
else if(ThisSum < 0)
ThisSum = 0;
}
return MaxSum;
}