int Max3( int A, int B, int C ) { /* Returns the maximum of 3 integers */ return A > B ? A > C ? A : C : B > C ? B : C; } int DivideAndConquer( int List[], int left, int right ) { /* Divide and conquer to find the maximum sum of sub columns from List[left] to List[right] */ int MaxLeftSum, MaxRightSum; /* Storing solutions of left and right subproblems */ int MaxLeftBorderSum, MaxRightBorderSum; /*Store results across boundaries*/ int LeftBorderSum, RightBorderSum; int center, i; if( left == right ) { /* The termination condition of recursion. The child column has only one number */ if( List[left] > 0 ) return List[left]; else return 0; } /* Here is the process of "dividing" */ center = ( left + right ) / 2; /* Find the midpoint */ /* Recursively finding the maximum sum of the two side sub columns */ MaxLeftSum = DivideAndConquer( List, left, center ); MaxRightSum = DivideAndConquer( List, center+1, right ); /* Find the sum of the largest subcolutions across the boundary line below */ MaxLeftBorderSum = 0; LeftBorderSum = 0; for( i=center; i>=left; i-- ) { /* Scan from midline to left */ LeftBorderSum += List[i]; if( LeftBorderSum > MaxLeftBorderSum ) MaxLeftBorderSum = LeftBorderSum; } /* End of left scan */ MaxRightBorderSum = 0; RightBorderSum = 0; for( i=center+1; i<=right; i++ ) { /* Scan right from midline */ RightBorderSum += List[i]; if( RightBorderSum > MaxRightBorderSum ) MaxRightBorderSum = RightBorderSum; } /* End of right scan */ /* Now return the result of "governance" */ return Max3( MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum ); } int MaxSubseqSum3( int List[], int N ) { /* Keep the same function interface as the first two algorithms */ return DivideAndConquer( List, 0, N-1 ); }

Application conditions of divide and conquer method:

1. The original problem can be divided into several solutions to the original problem;

2. The subproblem can be decomposed and solved;

3. The solution of the subproblem can be merged into the solution of the original problem;

4. The decomposed subproblems should be independent of each other, that is, they do not contain overlapping subproblems

MaxLeftSum, MaxRightSum; the maximum value of a single small block can be calculated (for example, 1, 2, -3,4, which can indicate that there is an independent maximum sub column 4 on the right)

MaxLeftBorderSum, MaxRightBorderSum; these two are responsible for finding the maximum value of the whole sequence

Let the total time complexity be T(N), divide it into two parts, each side is T(N/2), and add o (N) (it means that every element has been scanned once, i.e. the constant times of N), till N/=1, that is, the termination of recursion. The left and right sides are equal. T(1)=O(1)，k =logN, =N,ckN=cNlogN=O(NlogN)

When the two complexities are together, we take the larger O(NlogN),