Leetcode 33. 搜索旋转排序数组

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int beg = 0;
        int end = nums.size()-1;
        while(beg <= end)
        {
            int mid = beg + (end-beg)/2;
            //找到就直接返回
            if(target == nums[mid])
                return mid;

            else if( target < nums[mid])
            {
                if(nums[beg] < nums[mid])//[beg, mid]是递增区间 [mid+1, end]包含了旋转区间
                {
                    if(target >= nums[beg])  // 递增区间 直接用二分就可以了
                    {
                        end = mid -1;
                    }// 表明可能在后面的旋转区间里面
                    else
                    {
                        beg = mid +1;
                    }
                }
                else if(nums[beg] > nums[mid]) //[beg, mid]是旋转区间 [mid+1, end]包含了递增区间
                {
                    end = mid -1;
                }//前提是 target < nums[mid]  仅当还剩下两个数的时候，就相当于是target < nums[beg] 因此应该
                else if(nums[beg] == nums[mid])
                {
                    beg = mid+1;
                }
            }//
            else if(target > nums[mid])
            {
                if(nums[beg] < nums[mid])//[beg,mid]是递增区间，[mid+1, end]包含了旋转区间
                {
                    beg  = mid+1;
                }
                else if(nums[beg] > nums[mid])//前面是旋转区间，后面是递增区间，
                {
                    if(target >= nums[beg])//nums[beg] > nums[end]
                    {
                        end = mid -1;
                    }
                    else
                    {
                        beg = mid +1;
                    }
                }// 如果仅仅剩下两个元素了，target > nums[mid]
                else if(nums[beg] == nums[mid])
                {
                    beg = mid +1;
                }
            }
        }
        return -1;
    }
};

原文地址：https://www.cnblogs.com/randyniu/p/9354097.html