Recursion - Data Structures and Problem Solving Using Java - page 322

A recursive algorithm requires specifying a base case. When the size of
the problem reaches one element, we do not use recursion. The resulting Java
method is coded in Figure 7.20.

figure 7.20
A divide-and-conquer
algorithm for the
maximum contiguous
subsequence sum
problem

1 /**
2 * Recursive maximum contiguous subsequence sum algorithm.
3 * Finds maximum sum in subarray spanning a[left..right].
4 * Does not attempt to maintain actual best sequence.
5 */
6 private static int maxSumRec( int [ ] a, int left, int right )
7 {
8 int maxLeftBorderSum = 0, maxRightBorderSum = 0;
9 int leftBorderSum = 0, rightBorderSum = 0;
10 int center = ( left + right ) / 2;
11
12 if( left == right ) // Base case
13 return a[ left ] > 0 ? a[ left ] : 0;
14
15 int maxLeftSum = maxSumRec( a, left, center );
16 int maxRightSum = maxSumRec( a, center + 1, right );
17
18 for( int i = center; i >= left; i-- )
19 {
20 leftBorderSum += a[ i ];
21 if( leftBorderSum > maxLeftBorderSum )
22 maxLeftBorderSum = leftBorderSum;
23 }
24
25 for( int i = center + 1; i <= right; i++ )
26 {
27 rightBorderSum += a[ i ];
28 if( rightBorderSum > maxRightBorderSum )
29 maxRightBorderSum = rightBorderSum;
30 }
31
32 return max3( maxLeftSum, maxRightSum,
33 maxLeftBorderSum + maxRightBorderSum );
34 }
35
36 /**
37 * Driver for divide-and-conquer maximum contiguous
38 * subsequence sum algorithm.
39 */
40 public static int maxSubsequenceSum( int [ ] a )
41 {
42 return a.length > 0 ? maxSumRec( a, 0, a.length - 1 ) : 0;
43 }