用二分搜索来找到转折点，也就是最小数的位置。对二分搜索要稍作修改，当a[left]<=a[mid]，可以肯定a[left...mid]是升序的，那么a[left]是最小的，也可能最小的在a[mid+1...right]中，所以要比较a[left]和min{a[mid+1...right]}，同样对于a[left]>a[mid]，做相应的处理。

``` class Solution {
public:
int findPos(vector<int>& nums,int left,int right){
//if (nums.empty())//多余
//    return -1;
if (left > right||left>=nums.size())//后面这个条件不要，一些测试用例会产生下标越界错误
return -;
int mid = (left + right) / ;
if (nums[left] <= nums[mid])
{
int pos = findPos(nums, mid + , right);
if (pos == -)
return left;
else
return nums[left] < nums[pos] ? left : pos;
}
else
{
int pos = findPos(nums, left, mid - );
if (pos == -)
return mid;
else
return nums[mid] < nums[pos] ? mid :pos;
}
}
int bsearch(vector<int>nums, int left, int right, int key)
{
if (left>right)
return -;
int mid = (left + right) / ;
if (nums[mid] == key)
return mid;
if (nums[mid] < key)
return bsearch(nums, mid + , right, key);
else
return bsearch(nums, left, mid - , key);
}
int search(vector<int>& nums, int target)
{
int len = nums.size();
int pos = findPos(nums, , len);
int index;
if ((index=bsearch(nums, , pos - ,target)) != -)
return index;
else return bsearch(nums, pos, len, target);

}
};```

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