leetcode------Search in Rotated Sorted Array

标题：	Search in Rotated Sorted Array
通过率：	28.9%
难度：	难

Suppose a sorted array is rotated at 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 are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

这个题目是升级版的难度定为中等，我再做这个升级版时不知道什么旋转排序数组，然后来先做这个版本，然后我才弄明白，这种类型有两种情况

1.    4567012中间为7且   start=4<7,那么target如果在4-7之间的或者在7以外

2.45670123中间位置为0 且 start=4>0，那么target在4-0之间或者0以外。

直接看代码，最后一个else操作是start++  看一个例子 【1，3】target=3.当mid=0时 start=0 ，mid对应的元素和start对应的元素是相等，但是又不等于target

代码如下：

 1 public class Solution {

 2     public int search(int[] A, int target) {

 3         int start=0,end=A.length-1,mid=0;

 4         while(start<=end){

 5             mid=(start+end)/2;

 6             if(A[mid]==target)return mid;

 7             if(A[start]<A[mid]){

 8                 if((target<A[mid])&&(target>=A[start]))

 9                     end=mid-1;

10                     else start=mid+1;

11             }

12             else if(A[start]>A[mid]){

13                 if((target<A[mid])||(target>=A[start]))

14                     end=mid-1;

15                     else start=mid+1;

16             }

17             else start++;

18         }

19         return -1;

20     }

21 }

python：

 1 class Solution:

 2     # @param A, a list of integers

 3     # @param target, an integer to be searched

 4     # @return an integer

 5 

 6     def search(self, A, target):

 7         left = 0

 8         right = len(A) - 1

 9 

10         while left <= right:

11             mid = (left + right) >> 1

12             if target == A[mid]:

13                 return mid

14             # print mid, A[mid], left, A[left], right, A[right], '..',

15             if A[left] <= A[mid]:

16                 if A[left] <= target and target <= A[mid]:

17                     right = mid - 1

18                 else:

19                     left = mid + 1

20             else:

21                 if A[mid] < target and target <= A[right]:

22                     left = mid + 1

23                 else:

24                     right = mid - 1

25         return -1