Introduction to Algorithms (third edition) Exercises2.3

2.3-1:

3 9 26 38 41 49 52 59

3 26 41 52 9 38 49 57

3 41 52 26 38 57 9 49

3 41 52 26 38 57 9 49

2.3-2: (merge Sort)

voidMergeSort (intA[],intLintR) {    intm; if(L <r) {m= (L + r)/2;        MergeSort (A, L, M); MergeSort (A, M+1, R);    Merge (A, L, R, M); }}voidMergeintA[],intLintRintp) {    intI, j, K; intN1 = P-l +1; intN2 = R-p; intlarray[n1], rarray[n2];  for(i=0; i<n1; i++) Larray[i] = a[l+i]; //for (i=0; i<n1; i++) printf ("%d", Larray[i]);     for(j=0; j<n2; J + +) Rarray[j] = a[p+j+1]; //for (j=0; j<n2; j + +) printf ("%d", rarray[j]);i =0; J=0;  for(k=l; k<=r; k++)    {        if(I < N1 && (J >= N2 | | larray[i] <=Rarray[j])) {A[k]=Larray[i]; ++i; }        Else if(J < N2 && (i >= n1 | | rarray[j] <Larray[i])) {A[k]=Rarray[j]; ++J; }    }}

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2.3-3:

n=2:tn=2lg2=2

Assuming n=k, the equation is set.

tk+1=2 (Tk) + 2k+1=2 * (2k * k) + 2k+1=2k+1 (k+1)

2.3-4:

Worst case scenario:

1 n=2;

Tn = tn-1+n-1;

2.3-5: (Two-point search)

intBinarySearch (intA[],intVintN) {    intL, R, M; L=0; R= N-1;  while(L <=r) {m= (L + r)/2; if(v < a[m]) R = m-1; Else if(V > a[m]) L = m +1; Else returnm; }    return-1;}

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Worst case scenario:

2 n=2;

Tn=t (N/2) + 1 n=2k;

Θ (LGN)

2.3-6:

Unable to reduce run time

2.3-7: (Refer to the answer of netizens, do a bit of optimization, I hope you can make better suggestions for improvement)

Algorithm: asks if a set of n numbers contains and is 2 elements of X (need to support C99 standard)

BOOLSumX (intA[],intNintx) {    intlength, complement[n]; MergeSort (A,0, N-1); Length=deduplication (A, n);    Xcomplement (A, complement, length, x); returnMergesearch (A, complement, length);}BOOLMergesearch (intA[],intB[],intN) {    intI, J;  for(i=0, j=n-1; I<n && j>=0&& a[i]!=B[j];) if(A[i] < b[j]) i++; Elsej--; returnI<n && j>=0;}voidXcomplement (intA[],intAcomplement[],intNintx) {    inti;  for(i=0; i<n; i++) Acomplement[i] = x-a[i];}intDeduplication (intA[],intN) {    intI, tmp[n]; intj =0;  for(i=0; i<n; i++)    {        if(I >0&& a[i-1] = A[i]) J + +; TMP[J]=A[i]; }     for(i=0; i<=j; i++) A[i] =Tmp[i]; returnj+1;}voidMergeSort (intA[],intLintR) {    intm; if(L <r) {m= (L + r)/2;        MergeSort (A, L, M); MergeSort (A, M+1, R);    Merge (A, L, R, M); }}voidMergeintA[],intLintRintm) {    intMax = +; intI, j, K; intN1 = M-l +1; intN2 = R-m; intlarray[n1+1], rarray[n2+1];  for(i=0; i<n1; i++) Larray[i] = a[l+i];  for(j=0; j<n2; J + +) Rarray[j] = a[m+j+1]; LARRAY[N1]=Max; RARRAY[N2]=Max; I=0; J=0;  for(k=l; k<=r; k++)    {        if(I < N1 && Larray[i] <=Rarray[j]) {A[k]=Larray[i]; ++i; }        Else{A[k]=Rarray[j]; ++J; }    }}

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Θ (NLGN)

Introduction to Algorithms (third edition) Exercises2.3