The largest and most continuous sub-vectors in a one-dimensional vector

Q: A one-dimensional vector: arr[n] = {I1,i2,i3,......, in}, calculates the maximum and the number of its successive sub-vectors. (That is, the interception of successive paragraphs makes the elements and the largest, the element has a negative value; The sub-vector can be empty, i.e., and the minimum is 0)

A:

The initial idea was to be exhaustive, and the double loop would count all successive elements and

for [i,n) {    for[j,n) {        caculate sum (arr[j],arr[j]);    }   }

It works, but it looks stupid.

This assumes that the Intercept interval is arr[start, end] , then Arr[start, end+1] the maximum of the continuous sub-vector and can be determined by arr[start, end]?

If Arr[start, end] conforms to the condition of the sub-vector (assuming its maximum and Maxsofar) contains the last element, then Arr[start, end+1] only need to see (end+1) This position,

If Arr[start, end] Qualifying sub-vectors do not contain the last element,

Then Arr[start, End+1] The qualifying sub-vector is either arr[start, end], or the new sub-vector containing (end+1) this element (its maximum and that is, Arr[start,end] The maximum and maxending+arr[end+1] of the last element);

So the arr[n] has the following code:

#defineN 20#defineMAX (a) (a) > (b)? (a):(B)Static intMaxsofar=0; //Indicates the maximum and the continuous sub-vectors in the Arr[start, end] Static intmaxending=0; // represents the maximum and the (end) element of the Arr[start, end]  /*static int start =-1; * * The location of//start has not found a good way to determine Static intEnd =-1;
voidFindmax () {intI=0;  for(; i<n; + +i) { maxending = MAX (maxending+arr[i], 0 ); // if maxending plus arr[end+1] in arr[start,end] is less than 0, then maxending in arr[start,end+1] is 0;      Maxsofar = MAX (Maxending,maxsofar); Maxsofar in//arr[start,end+1] is the maximum value arr[start,end+1 in Maxsofar and maxending] in arr[start,end];               if(maxsofar==maxending) End=i; }}< Span style= "color: #800080;" >< Span style= "color: #800080;" >/ * When arr[n] = { -39,45,26,-18,-92,17,-4,68,72,-2,99,4,-67,142,55,-59,19,-41,-72,13}; The output is maxsofar=384,end=14; */            

The above-mentioned implementation is linear, and the efficiency is much higher than the N2 level of the poor lifting.

The largest and most continuous sub-vectors in a one-dimensional vector