To find a number in an ordered array of rotation (algorithm)

Topic:

Suppose an ordered array rotates at a certain position, and a new array is rotated, which is an ordered array of rotations. such as: (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2 ).

Now given an array of these, look for a number in the array. If found, returns the subscript, otherwise returns-1;

Ideas:

Idea 1:

Consider the characteristics of an ordered array of rotations: The front part is incremented, and the latter part is incremented, that is, the two parts are incrementing an ordered array, so you can use a binary search to do it.

Suppose that the array is a, the subscript is from left to right, and the number to find is target, then mid= (left+ ((right-left) >>1))

If A[mid]==target, return mid;

What if A[mid]>target or A[mid]<target? In mid, the left and right sides are not necessarily an ordered array. At this point we don't compare the size of A[mid] and target.

In mid, the left and right sides must have an ordered array of increments on one side, so this is the interval we're looking for.

If A[mid]>a[left], then the left part of mid is an ascending ordered array, if TARGET>=A[LEFT] && Target<a[mid], then right=mid-1; otherwise, left=mid +1;

If A[mid]<a[left], then the right part of mid is an ascending ordered array, if TARGET>A[MID] && target<=a[right], then left=mid+1; otherwise, right= Mid-1;

Idea 2:

Of course, we can also think from the normal thinking, from the A[mid]==target,a[mid]>target,a[mid]<target three parts to consider the boundary conditions. (See Code)

Code:

#include <iostream>using namespacestd;intRotatearraysearch_1 (int*a,intLeninttarget) {    intleft=0; intright=len-1;  while(left<=Right ) {        intmid=left+ (Right-left) >>1); if(a[mid]==target)returnmid; if(a[mid]>A[left]) {            if(Target>=a[left] && target<A[mid]) right=mid-1; Else Left=mid+1; }        Else{            if(Target>a[mid] && target<=A[right]) left=mid+1; Else Right=mid-1; }    }    return-1;}intRotatearraysearch_2 (int*a,intLeninttarget) {    intleft=0; intright=len-1;  while(left<=Right ) {        intmid=left+ (Right-left) >>1); if(a[mid]==target)returnmid; Else if(target>A[mid]) {            if(A[mid]>a[left] | | target<=A[right]) left=mid+1; Else Right=mid-1; }        Else{            if(A[mid]<a[right] | | target>=A[left]) right=mid-1; Else Left=mid+1; }    }    return-1;}intMain () {inta[]={2,3,4,5,1}; intb[]={5,1,2,3,4}; intn=sizeof(A)/sizeof(a[0]); cout<< rotatearraysearch_1 (A,n,5) <<Endl; cout<< rotatearraysearch_1 (A,n,3) <<Endl; cout<< rotatearraysearch_2 (B,n,1) <<Endl; cout<< rotatearraysearch_2 (B,n,4) <<Endl; return 0;}

To find a number in an ordered array of rotation (algorithm)