資料結構與演算法分析（C語言版）第二章習題2.7代碼實現

標籤：資料結構與演算法分析（c語言版）第二章習題2.7代碼實現

#include<stdio.h>#include<stdlib.h>#include<time.h>void swap(int *a,int *b){    if(*a!=*b)    {        /*         演算法1         *a+=*b;         *b=*a-*b;         *a-=*b;                  演算法2         *a*=*b;         *b=*a/*b;         *a/=*b;         演算法3         tmp=*a;         *a=*b;         *b=tmp;         演算法4*/        *a^=*b;        *b^=*a;        *a^=*b;    }}int Random(int i,int j){    if(i==0)        return rand()%(j+1);    else        return rand()%(j-i+1)+i;}void randSwap1(int A[],int n){    int i,j,rd;    srand(time(0));    for(i=0;i<n;i++)    {        rd=Random(1,n);        for(j=0;j<i;j++)        {            if(rd==A[j])            {                j=-1;                rd=Random(1,n);            }        }         A[i]=rd;      }}void randSwap2(int A[],int n){    int i,rd,*used;    used=(int*)malloc(n*sizeof(int));    for(i=0;i<n;i++)        used[i]=0;    srand(time(0));    for(i=0;i<n;i++)    {        rd=Random(1,n);        while(1)        {            if(used[rd-1])                rd=Random(1,n);            else                break;        }        A[i]=rd;         used[rd-1]=1;       }}void randSwap3(int A[],int n){    int i;    srand(time(0));    for(i=0;i<n;i++)        swap(A+i,A+Random(0,i));}int main(){    int i,n,*A;    printf("n=");    scanf("%d",&n);    A=(int *)malloc(n*sizeof(int));    randSwap1(A,n);    for(i=0;i<n;i++)        printf("%d ",A[i]);    printf("\n\n");    randSwap2(A,n);    for(i=0;i<n;i++)        printf("%d ",A[i]);    printf("\n\n");    randSwap3(A,n);    for(i=0;i<n;i++)        printf("%d ",A[i]);    printf("\n\n");    return 0;}

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資料結構與演算法分析（C語言版）第二章習題2.7代碼實現