最大子串和问题

question:

     Given an array that has positive and negative numbers, try to find a maximum subarray whose sum is the best.

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暴力解法:    复杂度(N^2)

pseudo-code:

    FIND-MAXIMUM-SUBARRAY(A, low, high)

1. sum = INT_MIN
2. for i = low to high - 1
3.     tmp_sum = A[i]
4.     for j = i + 1 to high
5.         tmp_sum = tmp_sum + A[j]
6.         if tmp_sum > sum
7.             sum = tmp_sum
8.             left = i
9.             right = j
10. return (left, right, sum)

typedef struct
{
       int low ;
       int high ;
       int sum ;
} Array_Info;

Array_Info Find_Maximum_Subarray_Force (int A[], int low , int high )
{
       Array_Info result ;
       int i , j ;
       int sum ;

       result .sum = INT_MIN ;

       for ( i = low; i < high ; i ++)
       {
             sum = A[ i ];
             for ( j = i + 1 ; j <= high ; j ++)
             {
                   sum += A[ j ];
                   if ( sum > result. sum )
                   {
                         result .low = i ;
                         result .high = j ;
                         result .sum = sum ;
                   }
             }
       }
       return result ;
}
// can not to solve the array that all the numbers are negitive

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分治法:    复杂度(NlogN)

pseudo-code:

    FIND-MAX-CROSSING-SUBARRAY(A, low, mid, high)

1. left-sum = INT_MIN
2. sum = 0
3. for i = mid downto low
4.     sum = sum + A[i]
5.     if sum > left-sum
6.         left-sum = sum
7.         max-left = i
8. right-sum = INT_MIN
9. sum = 0
10. for j = mind + 1 to high
11.     sum = sum + A[j]
12.     if sum > right-sum
13.         left-sum = sum
14.         max-right = j
15. return(max-left, max-right, left-sum + right-sum)

    FIND-MAXIMUM-SUBARRAY(A, low, high)

1. if low = high
2.     return (low, high, A[low])
3. else    mid = (low + high) / 2
4.     (left-low, left-high, left-sum) =
5.         FIND-MAXIMUM-SUBARRAY(A, low, mid)
6.     (right-low, right-high, right-sum) =
7.         FIND-MAXIMUM-SUBARRAY(A, mid + 1, high)
8.     (cross-low, cross-high, cross-sum) =
9.         FIND-MAX-CROSSING-SUBARRAY(A, low, mid, high)
10.     if left-sum >= right-sum and left-sum >= cross-sum
11.         return (left-low, left-high, left-sum)
12.     elseif right-sum >= left-sum and right-sum >= cross-sum
13.         return (right-low, right-high, right-sum)
14.     else return (cross-low, cross-high, cross-sum)

C语言实现:

typedef struct
{
       int low ;
       int high ;
       int sum ;
} Array_Info;

Array_Info Find_Max_Crossing_Subarray (int A[], int low , int mid , int high )
{
       int i , j ;
       int max_left , max_right ;
       int left_sum = INT_MIN ;
       int sum = 0 ;
       Array_Info cross ;

       for ( i = mid; i >= 0 ; i --)
       {
             sum += A[ i ];
             if ( sum > left_sum)
             {
                   left_sum = sum;
                   max_left = i;
             }
       }
       int right_sum = INT_MIN ;
       sum = 0;
       for ( j = mid + 1 ; j <= high ; j ++)
       {
             sum += A[ j ];
             if ( sum > right_sum)
             {
                   right_sum = sum;
                   max_right = j;
             }
       }

       cross .low = max_left ;
       cross .high = max_right ;
       cross .sum = left_sum + right_sum;
       return cross ;
}

Array_Info Find_Maximum_Subarray( int A [], int low, int high )
{
       Array_Info base , left , right , cross ;

       if ( low == high)
       {
             base .low = low ;
             base .high = high ;
             base .sum = A [low ];
             return base ;
       }

       int mid = ( low + high ) / 2;
       left = Find_Maximum_Subarray( A , low , mid );
       right = Find_Maximum_Subarray( A , mid + 1, high);
       cross = Find_Max_Crossing_Subarray (A , low , mid , high );

       if ( left .sum >= right .sum && left .sum >= cross .sum )
             return left ;
       else if (right . sum >= left . sum && right . sum >= cross . sum)
             return right ;
       else
             return cross ;
}

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线性解法：复杂度(N)

<1>Analyse

     从数组左边界开始，从左到右处理，记录到目前为止最大子数组的。若已知A[1....j]的最大子数组，基于如下性质将解扩展为A[1....j + 1]的最大子数组：

          A[1....j + 1]的最大子数组为 A[1....j] 的最大子数组或A[i....j + 1] (1 <= i <= j + 1)。

     性质分析：

          若A[j + 1] >= 0, 并且最大子数组为 A[i....j], 那么A[1....j + 1]最大子数组为A[i....j + 1];

          若A[j + 1] >= 0,若最大子数组为A[m....n] (1 <= m < n, n < j),那么 A[1....j + 1]最大子数组为A[1.....j]中最大子数组或A[i....j + 1], A[i...j + 1]满足不存在k (j > k > i)使A[i]+ A[i + 1]..+A[k] < 0

          若A[j + 1] < 0, 并且最大子数组为 A[m....n] (1 <=m < n, n <= j),那么A[1...j + 1]中最大子数组为A[m...n];

pseudo-code:

    FIND-MAXIMUM-SUBARRAY(A, low, high)

1. sum = INT_MIN
2. max = INT_MIN
3. for i = low to high
4.     if max < 0
5.         max = A[i]
6.     else max = max + A[i]
7.     if sum < max
8.         sum = max
9. return sum

c语言实现：

int Find_Maximum_Subarray ( int A [], int low , int high)
{
       int sum = INT_MIN ;
       int max = INT_MIN ;

       for ( int i = low ; i <= high; i++)
       {
             if ( max < 0)
                  max = A[ i ];
             else
                  max += A[ i ];

             if ( sum < max)
                   sum = max;
       }
       return sum ;
}

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备注：

     When all the numbers are negative, the function will return the minimun number of the array;