二分查找

方法1：递归

#include <iostream.h> #define SIZE 10 int data[SIZE] = {10,20,30,40,50,60,70,80,90,100}; int two_search(int start, int end, int val); main() { int val; int p; cin>>val; p=two_search(0, SIZE-1, val); if (p != -1) cout<<"Search"<<data[p]<<"is"<<p+1 <<" count!"; else cout<<"no found!"; return 0; } int two_search(int start, int end, int val) { int p = (start+end)/2; if(data[p]!=val) { if (start>end) return -1; if (data[p] > val) two_search(start,p-1,val); else two_search(p+1,end,val); } else return p; }

方法2：递归

#include<iostream> #define S 10 using namespace std; int binarysearch(int a[],int left,int right,int numtofind) { if(a[(left+right)/2]!=numtofind) { if(left > right) return -1; if(a[(left+right)/2]>numtofind) binarysearch(a,left,(right+left)/2- 1,numtofind); else binarysearch(a,(left+right)/2+1,right,numtofind); } else return (left+right)/2; } int main() { int a[S]={0,1,3,5,6,9,10,12,16,19}; int n; int pos; cin>>n; pos=binarysearch(a,0,S-1,n); cout<<"the num is on pos:"<<pos<<endl; }

3.循环

#include<iostream> using namespace std; int binary_search(int a[],int beg,int end,int num) { if (beg>end) return -1; if (num==a[(beg+end)/2]) return (beg+end)/2; else return num>a[(beg+end)/2]?binary_search(a,(beg +end)/2+1,end,num):binary_search(a,beg,(beg+end)/2-1,num); } int main() { int a[10]={1,2,3,4,5,7,8,9,10,11}; cout<<binary_search(a,0,9,6); }

4.循环

#include<iostream> #include<algorithm> using namespace std; int binarysearch(int a[],int left,int right,const int &value) { while(left<=right) { int middle=(left+right)/2; if(middle==value) return middle; else if(value>middle) left=middle+1; else right=middle-1; } return -1; } int main() { int a[10]={3,2,1,4,5,6,8,9,7,0}; sort(a,a+10); int n; cin>>n; if(binarysearch(a,0,9,n)==-1) cout<<"no"<<endl; else cout<<"yes"<<endl; }