【Leetcode】Search in Rotated Sorted Array II
从一个旋转的排序数组中寻找一个数字，数组中可能有重复数字，要求时间复杂度O(LogN)。
( 1 1 1 2 4 might become 1 1 2 4 1).
You are given a target value to search. If found in the array return true, otherwise return false.
思路：二分查找，需要注意的是，若中间数字跟最左端数字相等时，去掉最左端的数字，然后递归查找。代码如下：
class Solution
{
bool search(int A[], int n, int target)
{
return search(A,0,n-1,target);
}
bool search(int *ary, int start, int end, int target)
{
if (start > end) return false;
int mid = start + ((end-start)>>1);
int number0 = ary[start];
int number = ary[mid];
int number1 = ary[end];
if (number == target)
return true;
if (number > number0)
{
if(target < number0 || target > number)
return search(ary,mid+1,end,target);
else
return search(ary,start,mid-1,target);
}
else if(number < number0)
{
if (target > number1 || target < number)
return search(ary,start,mid-1,target);
else
return search(ary,mid+1,end,target);
}
// number == number0
return search(ary,start+1,end,target);
}
};