Maximum sub-sequence

 PackageALG;/*** Find the largest sub-series **/ Public classSubmaxarray { Public Static voidMain (string[] args) {int[] A =New int[] {1,-2, 3, 10,-4, 7, 2,-5 }; int[] B =New int[] {-6, 2, 4,-7, 5, 3, 2,-1, 6,-9, 10,-2 }; intMax =Maxsum (A, a.length);        SYSTEM.OUT.PRINTLN (max); intMAXB =Maxsum (b, b.length);                System.out.println (MAXB); LongWmax =MaxSubSum4 (a);        System.out.println (Wmax); LongMALXB =MaxSubSum4 (b);    System.out.println (MALXB); }    //the cycle of violence    Static intMaxsum (int[] A,intN) {intMaximum =Integer.min_value; intsum = 0;  for(inti = 0; I < n; i++) {             for(intj = i; J < N; J + +) {                 for(intK = i; K <= J; k++) {sum+=A[k]; }                if(Sum >maximum) {Maximum=sum; } Sum= 0;//remember to clear the zeros here, otherwise the sum of all the sub-arrays is finally stored.             }        }        returnmaximum; }    //Omit duplicate Calculations    Static intMaxsummodify (int[] a) {intMaxsum = 0;  for(inti = 0; i < a.length; i++) {            intThissum = 0;  for(intj = i; J < A.length; J + +) {thissum+=A[j]; if(Thissum >maxsum) Maxsum=thissum; }        }        returnmaxsum; }    /*** The largest sub-series of the divide-and- conquer algorithm, in the first half, in the second half or from the middle of the calculation **/    Static LongMaxsumrec (int[] A,intLeftintRight ) {        if(left = =Right ) {            if(A[left] > 0)                returnA[left]; Else                return0; }        intCenter = (left + right)/2; LongMaxleftsum =Maxsumrec (A, left, center); LongMaxrightsum = Maxsumrec (A, center + 1, right); //find the maximum value of the sequence ending with the last digit on the left        LongMaxleftbordersum = 0, leftbordersum = 0;  for(inti = center; I >= left; i--) {leftbordersum+=A[i]; if(Leftbordersum >maxleftbordersum) Maxleftbordersum=leftbordersum; }        //find the maximum value of the sequence ending with the next number on the right        LongMaxrightbordersum = 0, rightbordersum = 0;  for(intj = center + 1; J <= Right; J + +) {rightbordersum+=A[j]; if(Rightbordersum >maxrightbordersum) Maxrightbordersum=rightbordersum; }        returnmax3 (Maxleftsum, maxrightsum, Maxleftbordersum+maxrightbordersum); }    Static LongMAXSUBSUM3 (int[] a) {returnMaxsumrec (A, 0, a.length-1); }    Static LongMAX3 (LongALongBLongc) {if(A <b) {a=b; }        if(A >c)returnA; Else            returnC; }    //linear algorithm O (N)    /*** The basis of the algorithm: if A[i] is a negative number, then he cannot be the beginning of the maximal subsequence, the inference is that the maximum number of substrings of the beginning of the array and cannot be a negative number, * that is, a negative number of the subsequence can not be the beginning of the largest sub-sequence. * */    Static LongMAXSUBSUM4 (int[] a) {LongMaxsum = 0, thissum = 0;  for(intj = 0; J < A.length; J + +) {thissum+=A[j]; if(Thissum >maxsum) Maxsum=thissum; Else if(Thissum < 0) Thissum= 0; }        returnmaxsum; }    /*** If all negative, the largest subsequence is 0 for one element, * or the largest negative number? * */    Static intMaxsum (int[] a) {intmax = a[0];//full negative condition, return maximum number        intsum = 0;  for(intj = 0; J < A.length; J + +) {            if(Sum >= 0) {//If you add an element, sum>=0, addSum + =A[j]; }Else{sum= A[j];//If you add an element to the sum<0, you don't add            }            if(Sum >max) Max=sum; }        returnMax; }}

Maximum sub-sequence