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;

}