LeetCode 33. Search in Rotated Sorted Array

1. 题目描述
   Suppose an array sorted in ascending order 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.
   Your algorithm’s runtime complexity must be in the order of O(log n).
2. 解决方案
   可能解决本题目的方案有多种，这里作者提供两种解法。
   1）最常规的思路是采用折半查找法，但是由于原来的有序数组按照某个轴进行了宣战，所以，在判断条件上会有一些麻烦。如果数组没有按照某个轴旋转，则就是普通的这般查找；如果数组按照某个轴完成了旋转，则有nums[left] >= nums[right]，这是需要判nums[middle]、target、（nums[right]或nums[left]之间选择一个）这三者之间的大小关系，来更新left或right。

2. 第二种方法是找到旋转轴，然后根据旋转轴将旋转后的数组与原来的数组完成位置映射，从而利用传统的折半查找法即可解决。

3. 代码
   1）解决方案1代码

#include 
#include 
using namespace std;

int search(vector& nums, int target);

int main()
{
    vector vec;
    int a[] = {4,5,6,7,0,1,2};
    for (int i=0; i& nums, int target){
    if (nums.empty()){
        return -1;
    }
    int left = 0;
    int right = nums.size()-1;
    while(left <= right){
        int middle = (left + right) / 2;
        if (nums[left] < nums[right]){
            if (nums[middle] == target){
                return middle;
            }else if (nums[middle] > target){
                right = middle - 1;
            }else{
                left = middle + 1;
            }
        }else{
            if (nums[middle] == target){
                return middle;
            }else if (nums[middle] > target){
                if (nums[middle] <= nums[right]){
                    right = middle - 1;
                }else{
                    if (target <= nums[right]){
                        left = middle + 1;
                    }else{
                        right = middle - 1;
                    }
                }
            }else{
                if (nums[middle] <= nums[right]){
                    if (target > nums[right]){
                        right = middle - 1;
                    }else{
                        left = middle + 1;
                    }
                }else {
                    left = middle + 1;
                }
            }
        }
    }
    return -1;
}

2）解决方案二代码

#include 
#include 
using namespace std;

int search(vector& nums, int target);

int main()
{
    int a[] = {4,5,6,7,0,1,2};
    vector v;
    for (int i=0; i& nums, int target){
    int left = 0;
    int right = nums.size()-1;
    while(left < right){
        int mid = (left + right) / 2;
        if (nums[mid] > nums[right]){
            left = mid+1;
        }else{
            right=mid;
        }
    }
    int pivot = left;
    left = 0;
    right = nums.size() - 1;
    while(left <= right){
        int mid = (left + right) / 2;
        int newmid = (mid + pivot) % nums.size();
        if (target == nums[newmid]){
            return newmid;
        }else if (target > nums[newmid]){
            left = mid+1;
        }else{
            right = mid-1;
        }
    }
    return -1;
}