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;
    }
};