2.2.22三向歸并排序。假設每次我們是把數組分成三個部分而不是兩個部分并將它們分別排序，然后進行三向歸并。這種算法的運行時間的增長數量級是多少？
運算時間增長數量級是O(Nlg3N)。
public class E2d2d22
{
??? public static void sort(Comparable[] a)
??? {
??????? Comparable[] aux=new Comparable[a.length];
??????? sort(a,aux,0,a.length-1);
?? }
???
??? private static void sort(Comparable[] a,Comparable[] aux,int lo,int hi)
??? {
???????? if (hi<=lo)??? return ;
????????? int m1=lo+(hi-lo)/3;
????????? int m2=m1+1+(hi-m1-1)/2;
????????? sort(a,aux,lo,m1);
????????? sort(a,aux,m1+1,m2);
????????? sort(a,aux,m2+1,hi);
????????? merge(a,aux,lo,m1,m2,hi);
??? }
???
??? private static void merge(Comparable[] a,Comparable[] aux,int lo,int m1,int m2,int hi)
??? {
????? for(int k=lo;k<=hi;k++)
????????? aux[k]=a[k];
????? //
????? int aIndex=lo;
????? int bIndex=m1+1;
????? int cIndex=m2+1;
?????
????? Comparable A,B,C;
????? Double maxValue=Double.MAX_VALUE;
????? for(int k=lo;k<=hi;k++)
????? {
?????????? if (aIndex<=m1) A=aux[aIndex]; else A=maxValue;
?????????? if (bIndex<=m2) B=aux[bIndex]; else B=maxValue;
?????????? if (cIndex<=hi)?? C=aux[cIndex]; else C=maxValue;
?????????? //
?????????? if(less(A,B) && less(A,C))
???????????? {a[k]=A;aIndex++;}
?????????? else if(less(B,A) && less(B,C))
???????????? {a[k]=B;bIndex++;}
?????????? else if(less(C,A) && less(C,B))
???????????? {a[k]=C;cIndex++;}
?????????? else
?????????????? {a[k]=A;aIndex++;}
????? }
??? }
?
??? private static boolean less(Comparable v,Comparable w)
??? { return v.compareTo(w)<0;}
???
??? private static void exch(Comparable[] a,int i,int j)
??? {
????? Comparable t=a[i];
????? a[i]=a[j];
????? a[j]=t;
??? }
??
?public static boolean isSorted(Comparable[] a)
??? {
????? for(int i=1;i<a.length;i++)
??????? if(less(a[i],a[i-1])) return false;
????? return true;
??? }
?
?public static void main(String[] args)
?? {
???? Integer N=Integer.parseInt(args[0]);
???? Comparable[] a=new Comparable[N];
???? for(int i=0;i<N;i++)
???????? a[i]=StdRandom.uniform();
????
???? sort(a);
???? StdOut.printf("%s",isSorted(a));
? }
}

轉載于:https://www.cnblogs.com/longjin2018/p/9860119.html

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