LeetCode_Search in Rotated Sorted Array

题目：

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.

三个招：前两个很简单，巧妙地是第三个。

//方法一：分批次二分查找 ；查找出分割点需要时间O(n),总时间复杂度 O(n) 
int BinaryResearch(int A[],int low,int high,int target)
{
    int l = low;
    int h = high;
    while(l<=h)
    {
       int mid = (int)((h+l)/2);
       if(A[mid]==target) return mid;
       else if(A[mid]<target) l = mid+1;
       else h = mid-1;
    }
    return A[h]; 
}
int search1(int A[], int n, int target) {
   
     int index = 0;
     for(int i = 0;i<n-1;i++)
     {
       if(A[i]>A[i+1])
       {
         index = i;
         break;
       }
     }
     int a = BinaryResearch(A,0,index,target);
     int b = BinaryResearch(A,index+1,n-1,target);
     if(a==-1&&b==-1)
     return -1;
     else 
     return a==-1?b:a;
}
int search2(int A[], int n, int target) {
     //顺序查找 ,O(n)
     int index = -1;
     for(int i = 0;i<n;i++)
     {
     if(A[i]==target)
     {
        index = i;
     }}
     return index;
}
//完全的二分查找，O(logn) 
int search3(int A[], int n, int target) {
     int left = 0;
     int right = n-1;
     while(left<=right)
     {
            int mid = (int)((left + right)/2);
            if(A[mid] == target) return mid;
            if(A[left]<A[mid])//A[mid]在前半部分
            {
               if(target<A[mid]&&target>=A[left])
               right = mid-1;
               else left = mid+1; 
            } 
            else if(A[left]>A[mid])//A[mid]位于后半段 
            {
               if(target>A[mid]&&target<=A[right])
               left = mid+1;
               else 
               right = mid-1;
            }
            else left++; 
     }
     return -1;
     
}