"Leetcode" Search in rotated Sorted Array

Search in rotated Sorted Array

Suppose a sorted array is rotated on some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2 ).

You is given a target value to search. If found in the array is return its index, otherwise return-1.

Assume no duplicate exists in the array.

Using the dichotomy, if LEFT<A[MID], the description from left to right start to mid is ordered if LEFT>A[MID], the description of the rights start from mid, and to that is ordered if LEFT=A[MID], Then the left and mid are in the same position, if A[mid]!=target is the next left=left+1;

1 classSolution {2  Public:3     intSearchintA[],intNinttarget) {4        5         intleft=0;6         intright=n-1;7         intmid;8          while(left<=Right )9         {TenMid= (left+right)/2; One             A             if(A[mid]==target)returnmid; -             -             if(A[left]<a[mid])// Left the             { -                 if(a[left]<=target&&target<A[mid]) -                 { -right=mid-1; +                 } -                 Else +                 { ALeft=mid+1; at                 } -             } -             Else if(A[left]>a[mid])// Right -             { -                 if(a[mid]<target&&target<=A[right]) -                 { inLeft=mid+1; -                 } to                 Else +                 { -right=mid-1; the                 } *             } $             ElsePanax Notoginseng             { -                 //The Last Judgment statement can be put together in the first statement, if (A[left]<=a[mid]) theLeft=mid+1; +             } A             the         } +         return-1; -     } $};

"Leetcode" Search in rotated Sorted Array