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;     }