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;
}
};