Maximum sub-segments and problems

Maximum sub-segments and issues:

A sequence of n integers (possibly negative numbers) a[1],a[2],a[3],..., A[n], and the maximum value of the sub-segments of the sequence, such as A[i]+a[i+1]+...+a[j]. When the given integer is negative, the defined sub-segment and is 0, according to this definition, the optimal value is:

Max{0,a[i]+a[i+1]+...+a[j]},1<=i<=j<=n
For example, when (a1,a2,a3,a4,a4,a6) = ( -2,11,-4,13,-5,-2), the maximum child segment and is 20.

1. You can use a simple algorithm 3 for loop to calculate the answer, the time complexity is O (n^3).

2. The problem can be broken down by the use of a divide-and-conquer approach.

Divide a[1n] into A[1N/2] and a[n/2+1n], the maximum field of a[1n] and there are three cases:
(1) Maximum sub-segments of a[1n] and the largest sub-segments and the same as A[1N/2]
(2) Maximum sub-segments of a[1n] and the largest sub-segments and the same as a[n/2n]
(3) Maximum sub-segments of a[1n] and for Ai++aj,1<=i<=n/2,n/2+1<=j<=n

The time complexity is O (NLOGN).

3. Dynamic Programming Method:

B[j]=max{a[i]++a[j]},1<=i<=j, and 1<=j<=n, the maximum number of sub-segments to be asked and for Max B[j],1<=j<=n.
by b[j] The definition is easy to know, when b[j-1]>0 b[j]=b[j-1]+a[j], otherwise b[j]=a[j]. So B[J] 's dynamic programming recursion is:
B[j]=max (B[j-1]+a[j],a[j]), 1<=j<=n.

Time Complexity of O (n)

This gives the code

 Public Static intMaxsum (int[] a) {intn =a.length; intsum = 0, b = 0;  for(inti=0; i<n; i++) {            if(b > 0) {b+=A[i]; }            Else{b=A[i]; } Sum= B>sum?b:sum; }                returnsum;
}

Test data:

 Public Static void Main (string[] args) {        int[] A = {-2, one, -4, -5, 2};                 System.out.println (Maxsum (a));

Results:

  20

　　

Maximum sub-segments and problems