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：

public class Solution {
    public int search(int[] A, int target) {
        int start=0,end=A.length-1,mid=0;
        while(start<=end){
            mid=(start+end)/2;
            if(A[mid]==target)return mid;
            if(A[start]<A[mid]){
                if((target<A[mid])&&(target>=A[start]))
                    end=mid-1;
                    else start=mid+1;
            }
            else if(A[start]>A[mid]){
                if((target<A[mid])||(target>=A[start]))
                    end=mid-1;
                    else start=mid+1;
            }
            else start++;
        }
        return -1;
    }
}

python：

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer

    def search(self, A, target):
        left = 0
        right = len(A) - 1

        while left <= right:
            mid = (left + right) >> 1
            if target == A[mid]:
                return mid
            # print mid, A[mid], left, A[left], right, A[right], '..',
            if A[left] <= A[mid]:
                if A[left] <= target and target <= A[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                if A[mid] < target and target <= A[right]:
                    left = mid + 1
                else:
                    right = mid - 1
        return -1