问题及代码:
验证折半查找算法。请用有序表{12,18,24,35,47,50,62,83,90,115,134}作为测试序列,分别对查找90、47、100进行测试。
普通折半法:
#include <stdio.h> #define MAXL 100 typedef int KeyType; typedef char InfoType[10]; typedef struct { KeyType key; //KeyType为关键字的数据类型 InfoType data; //其他数据 } NodeType; typedef NodeType SeqList[MAXL]; //顺序表类型 int BinSearch(SeqList R,int n,KeyType k) { int low=0,high=n-1,mid; while (low<=high) { mid=(low+high)/2; if (R[mid].key==k) //查找成功返回 return mid+1; if (R[mid].key>k) //继续在R[low..mid-1]中查找 high=mid-1; else low=mid+1; //继续在R[mid+1..high]中查找 } return 0; } int main() { int i,n=10; int result; SeqList R; KeyType a[]= {12,18,24,35,47,50,62,83,90,115,134},x=90; for (i=0; i<n; i++) R[i].key=a[i]; result = BinSearch(R,n,x); if(result>0) printf("序列中第 %d 个是 %d\n",result, x); else printf("木有找到!\n"); return 0; }递归的折半查找算法
#include <stdio.h> #define MAXL 100 typedef int KeyType; typedef char InfoType[10]; typedef struct { KeyType key; //KeyType为关键字的数据类型 InfoType data; //其他数据 } NodeType; typedef NodeType SeqList[MAXL]; //顺序表类型 int BinSearch1(SeqList R,int low,int high,KeyType k) { int mid; if (low<=high) //查找区间存在一个及以上元素 { mid=(low+high)/2; //求中间位置 if (R[mid].key==k) //查找成功返回其逻辑序号mid+1 return mid+1; if (R[mid].key>k) //在R[low..mid-1]中递归查找 BinSearch1(R,low,mid-1,k); else //在R[mid+1..high]中递归查找 BinSearch1(R,mid+1,high,k); } else return 0; } int main() { int i,n=10; int result; SeqList R; KeyType a[]= {12,18,24,35,47,50,62,83,90,115,134},x=90; for (i=0; i<n; i++) R[i].key=a[i]; result = BinSearch1(R,0,n-1,x); if(result>0) printf("序列中第 %d 个是 %d\n",result, x); else printf("木有找到!\n"); return 0; }
输出及结果;
分析:
折半算法思想即通过对半的方法,节省了大量时间