Dynamic programming-the largest and most continuous sub-sequences

given the sequence of k integers {n1,n2,..., Nk}, any contiguous subsequence can be expressed as {Ni, ni+1, ..., Nj}, where 1 <= i <= j <= K. The maximal contiguous subsequence is the element and the largest of all successive sub-sequences, such as the given sequence {-2, 11,-4, 13,-5,-2}, whose maximum contiguous subsequence is {11,-4,13}, the maximum contiguous subsequence, and 20.

Note: For convenience, if all integers are negative, the maximum subsequence and is 0.

Algorithm one, the exhaustive method, finds all the sub-arrays, and then finds the and of the sub-array, and takes the maximum value in all the sub-arrays

/*o (n^3) Poor lift method * Disadvantage: Repeat accumulation, and maxsum comparison, each i->j in the middle of the total accumulated before compared with the maxsum * */public    static int MaxSubSequence1 (int[] array) {          int length=array.length;      int maxsum=0;      int thissum=0;      for (int i=0;i<length;i++) {for      (int j=i;j<length;j++) {      thissum=0;      for (int k=i;k<j;k++) {//I->j add up between the      thissum+=array[k];      if (thissum>maxsum) {      maxsum=thissum;            }}}} return maxsum;          }

Algorithm two, the first method of each i->j between each iteration, repeated calculation of a lot, you can take advantage of the computed sub-array and

/*o (n^2) Exhaustive method     * I->j between each summation and maxsum comparison     * */public    static int MaxSubSequence2 (int[] array) {        int Length=array.length;      int maxsum=0;      int thissum=0;      for (int i=0;i<length;i++) {     thissum=0;    for (int j=i;j<length;j++) {   thissum+=array[j];      if (thissum>maxsum) {  maxsum=thissum;}}       }       return maxsum;    }

algorithm three, dynamic programming, Initializes a maximum array of maxsum[n],Maxsum[i] is the maximum sum of the subarray that represents a[0...i] ending with a[i], then maxsum[i] is equal to a[0...i-1] maximum and plus a[i] in and A[i] compare Max, maxsum[i] = max {Maxsum[i-1] + a[i], A[i]}

  Dynamic programming, state equation maxsum[i] = max{maxsum[i-1] + a[i], a[i]}; Maxsum[i] indicates the largest and public    static int MaxSubSequence3 (int[] array) with the end of A[i] {         int length=array.length;    Int[] Maxsum=new int[length];    MAXSUM[0]=ARRAY[0];      for (int i=1;i<length;i++) {      Maxsum[i]=math.max (maxsum[i-1]+array[i], array[i]);      }      Find the maximum value in maxsum      int maxsum=integer.min_value;      for (int i=0;i<maxsum.length;i++) {      if (maxsum[i]>maxsum) {      maxsum=maxsum[i];      }      }         return maxsum;    }

Improvement of Algorithm three O (n)

     /* Simplified method Three, when the previous summation and thissum less than 0 is set 0, discard, greater than maxsum, assign the value to maxsum     * */public    static int MaxSubSequence4 (int[] Array) {        int length=array.length;      int maxsum=0;      int thissum=0;      for (int i=0;i<length;i++) {      thissum+=array[i];      if (thissum>maxsum) {      maxsum=thissum;      }      else if (thissum<0) {      thissum=0;      }      }        return maxsum;    }

Dynamic programming-the largest and most continuous sub-sequences