33 - Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

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思路分析：

此时为乱序的情况

class Solution {
public:
    int se(int st, int ed, int target, vector<int>& nums)
    {
        if (st > ed)
        {
            return -1;
        }
        else
        {
            int mid = st + (ed - st)/2;
            
            if (nums[mid] == target)
            {
                return mid;
            }
            
            if (nums[mid] >= nums[st])
            {
                if (target <= nums[mid] && target >= nums[st])
                {
                    return se(st, mid - 1, target, nums);
                }
                else
                {
                    return se(mid + 1, ed, target, nums);
                }
                
            }
            
            if (nums[mid] < nums[st])
            {
                if (target <= nums[mid] || target >= nums[st])
                {
                    return se(st,mid-1,target,nums);
                }
                else
                {
                    return se(mid+1,ed,target,nums);
                }
            }
        }
        
        return -1;
        
    }

    int search(vector<int>& nums, int target) 
    {
        if (nums.size() == 0)
        {
            return -1;
        }
        
        return se(0, nums.size() - 1, target, nums);
        
    }
};