The art of algorithms-maximum subsequences and Problems

 

Problem description: There is a sequence like 23,-23,11, 43,-45, 29, 34,0, 23,-12 to find the sum of the largest subsequences in this sequence, for example, from 0th to 3rd elements are a sub-sequence with the sum of 54. The shortest sub-sequence can have only one element, and the longest sub-sequence can contain all elements.

 

The following is an algorithm for solving this problem (implemented in Java ):

 

Algorithm 1 is the first algorithm we think of. It is a very easy-to-understand algorithm, but it has the lowest efficiency. The average time complexity is O (n ^ 3 ):

 

 

Public static int getMaxSubVector1 (int [] m ){

Int maxSubVector = 0;

Int I, j = 0, k = 0;

For (I = 0; I <m. length; I ++ ){

For (j = I; j <m. length; j ++ ){

Int sum = 0;

For (k = I; k <j; k ++ ){

Sum + = m [k];

MaxSubVector = Math. max (maxSubVector, sum );

}

}

}

Return maxSubVector;

}

Algorithm 2, a slightly improved algorithm with an average time complexity of O (n ^ 2)

 

 

Public static int getMaxSubVector2 (int [] m ){

Int maxSubVector = 0;

Int I, j = 0;

For (I = 0; I <m. length; I ++ ){

Int sum = 0;

For (j = I; j <m. length; j ++ ){

Sum + = m [j];

MaxSubVector = Math. max (maxSubVector, sum );

}

}

Return maxSubVector;

}

Algorithm 3, we can solve this problem using the idea of the divide and conquer algorithm, so that we can reduce the average time complexity to O (nlogn ):

 

 

Public static int getMaxSubVector4 (int [] B, int l, int u ){

Int sum = 0;

Int m = (l + u)/2;

If (l> u) return 0;

If (l = u) return Math. max (0, B [1]);

Int lmax = sum = 0;

For (int I = m; I> = 1; I --){

Sum + = B [I];

Lmax = Math. max (lmax, sum );

}

Int rmax = sum = 0;

For (int I = u; I> m; I --){

Sum + = B [I];

Rmax = Math. max (rmax, sum );

}

Return max3 (lmax + rmax, getMaxSubVector4 (B, l, m), getMaxSubVector4 (B, m + 1, u ));

}

Public static int max3 (int x, int y, int z ){

If (x <y ){

X = y;

}

If (x> z ){

Return x;

}

Return z;

}

Algorithm 4: this algorithm is a scanning idea and a linear time O (n ):

 

 

Public static int getMaxSubVector5 (int [] B ){

Int maxSubVector = 0;

Int maxEnding = 0;

For (int I = 0; I <B. length; I ++ ){

MaxEnding = Math. max (maxEnding + B [I], 0 );

MaxSubVector = Math. max (maxSubVector, maxEnding );

}

Return maxSubVector;

}

 

NOTE: For the above algorithm ideas, refer to the second edition of programming Pearl River