Leetcode "Search in rotated Sorted Array" Python implementation

title :

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

code : OJ Test via runtime:53 ms

1 classSolution:2     #@param A, a list of integers3     #@param target, an integer to be searched4     #@return An integer5     defSearch (self, A, target):6         #None Case & Zero case7         ifA isNoneorLen (A) = =0:8             return-19         #binary SearchTenStart =0 Oneend = Len (A)-1 A          whilestart<=End: -             #One element left case -             ifStart = =End: the                 ifa[start]==Target: -                     returnStart -                 Else: -                     return-1 +             #Elements left case -             ifStart+1 = =End: +                 ifa[start]==Target: A                     returnStart at                 elifa[end]==Target: -                     returnEnd -                 Else: -                     return-1 -             #equal or more than three elements case -Mid = (start+end)/2 in             ifa[mid]==Target: -                 returnMid to             elifA[mid]>Target: +                 ifA[start]>a[mid] anda[end]<A[mid]: -Start = Mid+1 the                 elifA[start]<a[mid] anda[end]<A[mid]: *                     ifa[end]>=Target: $Start = Mid+1Panax Notoginseng                     Else: -End = Mid-1 the                 elifA[start]>a[mid] andA[end]>A[mid]: +End = Mid-1 A                 Else: theEnd = Mid-1 +             Else: -                 ifA[start]>a[mid] anda[end]<A[mid]: $End = Mid-1 $                 elifA[start]<a[mid] anda[end]<A[mid]: -Start = Mid+1 -                 elifA[start]>a[mid] andA[end]>A[mid]: the                     ifa[end]>=Target: -Start = Mid+1Wuyi                     Else: theEnd = Mid-1 -                 Else: WuStart = Mid+1 -         return-1

Ideas :

This is the idea of binary search.

The individual did not think out what a good way to bite the bullet hard to write a violent solution.

The traditional binary search only needs to judge A[mid] and target size is OK, but this problem is rotated array, light judgment A[mid] is not enough.

It is also necessary to determine the size of A[start] a[end] and A[mid] in order to determine whether the target may fall within the [Start,mid] or [mid,end] interval.

Your code is a little cumbersome and ugly, and it is estimated that some if else conditions can be merged and subsequently improved.

Leetcode "Search in rotated Sorted Array" Python implementation