旋转有序数组的搜索

1、（无重复）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.

class Solution {
public:
int search(int A[], int n, int target) 
{
        if(A==NULL||n==0)return -1;
        int left=0,right=n-1;
        int mid;
        while(left<=right)
        {
            mid=left+(right-left)/2;
            if(A[mid]==target)return mid;
/*          if(A[mid]==A[left]&&A[mid]==A[right])
            {
                left++;
                right--;
            }
            else*/ if(A[mid]>=A[left])
            {
                if(A[mid]<target)left=mid+1;
                else
                {
                    if(A[left]<=target)right=mid-1;
                    else left=mid+1;
                }
            }
            else
            {
               if(A[mid]>target)right=mid-1;
               else
               {
                   if(A[right]>=target)left=mid+1;
                   else right=mid-1;
               }
            }
        }
        return -1;
}
};

　　

2（有重复）Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

class Solution {
public:
    bool search(int A[], int n, int target) {
        if(A==NULL||n==0)return false;
        int left=0,right=n-1;
        int mid;
        while(left<=right)
        {
            mid=left+(right-left)/2;
            if(A[mid]==target)return true;
            if(A[mid]==A[left]&&A[mid]==A[right])
            {
                left++;
                right--;
            }
            else if(A[mid]>=A[left])
            {
                if(A[mid]<target)left=mid+1;
                else
                {
                    if(A[left]<=target)right=mid-1;
                    else left=mid+1;
                }
            }
            else
            {
               if(A[mid]>target)right=mid-1;
               else
               {
                   if(A[right]>=target)left=mid+1;
                   else right=mid-1;
               }
            }
        }
        return false;
    }
};