The smallest number of offer-rotating arrays

Topic:

Moves the first element of an array to the end of the array, which we call the rotation of the array. Enter a rotation of an incrementally sorted array, outputting the smallest element of the rotated array.

For example, a rotation of the array (3,4,5,1,2) bit {1,2,3,4,5}, the minimum value of the array is 1.

Basic idea:

Binary lookup, P1 point A[0],p2 point to a[len-1]. If A[MID]>A[P1], the minimum value is in the second half, p1=mid; if a[mid]<a[p2], the minimum value is in the first half, P2=mid.

Special cases:

Note: In both arrays, the first number, the last number, and the median number are 1, and we cannot determine whether the median number 1 belongs to the first ascending word group or the second incrementing word group.

This situation can only be found in sequential methods.

#include <iostream>using namespace Std;int mininorder (int a[],int index1,int index2) {int result = A[index1];for ( int i=index1+1;i<=index2;i++) {if (Result>a[i]) result=a[i];} return result;} int min (int a[],int len) {if (len<0) return 0;int index1=0;int index2=len-1;int indexmid=index1;//If the rotated array is itself, The default is Index1while (A[index1]>=a[index2]) {if (index2-index1==1) {indexmid=index2;break;} Indexmid= (INDEX1+INDEX2)/2;//if the Index1,index2 and indexmid three are equal, you can only search sequentially if (a[index1]==a[index2]&&a[indexmid]= =A[INDEX1]) return Mininorder (A,INDEX1,INDEX2), if (A[indexmid]>=a[index1]) Index1=indexmid;else if (A[indexMid] <=A[INDEX2]) Index2=indexmid;} return a[indexmid];} int main () {int a[]={3,4,5,1,2};int b[]={1,0,1,1,1};cout<<min (a,5) <<endl;cout<<min (b,5) << Endl;return 0;}

Sword refers to the smallest number of offer-rotated arrays