以前总觉得二分查找简单，刷了一些题目后再也不觉得二分查找简单了，各种边界条件，简直要了老命，但对这道题而言，我们只要找到某半段有序数组即可，如果target恰好位于其中，那么问题就转换为普通的二分查找。下面是两种形式的写法：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0,r = nums.size() - 1;
        while(l <= r)
        {
            int mid = l + (r - l)/2;
            if(nums[mid] == target) return mid;
            if(nums[mid] < nums[r])
            {
                //后半段有序
                if(nums[mid] < target && target <= nums[r])
                    l = mid + 1;
                else
                    r = mid - 1;
            }else{
                //前半段有序
                if(nums[l] <= target && target < nums[mid])
                    r = mid - 1;
                else
                    l = mid + 1;
            }
        }
        
        return -1;
    }
};

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0,r = nums.size() - 1;
        while(l <= r)
        {
            int mid = l + (r - l)/2;
            if(nums[mid] == target) return mid;
            if(nums[mid] > nums[r])
            {
                //前半段有序,且target位于前半段
                if(nums[l] <= target && target < nums[mid])
                    r = mid - 1;
                else
                    l = mid + 1;
            }else{
                //后半段有序,且target位于后半段
                if(nums[mid] < target && target <= nums[r])
                    l = mid + 1;
                else
                    r = mid - 1;
            }
        }
        return -1;
    }
};

一种更好理解的思路
2021.06.10更新另一种思路：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0,r = nums.size() -1;
        while(l <= r)
        {
            int mid = l + (r-l)/2;
            if(nums[mid] == target) return mid;
            //[3,4,5,1,2]
            if(nums[mid] > nums[r])
            {
                if(target < nums[mid] && target >= nums[l])
                    r = mid - 1;
                else
                    l = mid + 1;
            //[5,1,2,3,4]
            }else if(nums[mid] < nums[l]){
                if(target > nums[mid] && target <= nums[r])
                    l = mid + 1;
                else
                    r = mid - 1;
            }else{
                if(target < nums[mid])
                    r = mid -1;
                else
                    l = mid + 1;
            }
        }
        return -1;
    }
};

仔细分析上面两种代码解法，实际是上是一样的。