33. Search in Rotated Sorted Array
Problem's Link
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Mean:
给你一个数组，这个数组是由两个有序的数组拼接成的（无重复元素），在这个数组中查找元素k的下标.
analyse:
既然是由两个有序数组拼接而成，那么只需找到断点，然后进行两次二分查找就行.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-01-21.22
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
class Solution
{
public :
int search( vector < int >& nums , int target)
{
int len = nums . size();
int low1 = 0 , high1 = len - 1 , low2 = 0 , high2 = len - 1;
for( int i = 1; i < len; ++ i)
{
if( nums [ i ] < nums [ i - 1 ])
{
high1 = i - 1 , low2 = i;
break;
}
}
if( high1 == len - 1 && low2 == 0)
return isExist( nums , low1 , high2 , target);
else
{
int idx1 = isExist( nums , low1 , high1 , target);
int idx2 = isExist( nums , low2 , high2 , target);
if( idx1 ==- 1 && idx2 ==- 1)
return - 1;
else
return idx1 !=- 1 ? idx1: idx2;
}
}
int isExist( vector < int >& nums , int low , int high , int target)
{
while( low <= high)
{
int mid =( low + high) >> 1;
if( target < nums [ mid ])
high = mid - 1;
else if( target > nums [ mid ])
low = mid + 1;
else return mid;
}
return - 1;
}
};
int main()
{
Solution solution;
int n , k;
vector < int > ve;
while( cin >>n >> k)
{
int tempVal;
for( int i = 0; i <n; ++ i)
{
cin >> tempVal;
ve . push_back( tempVal);
}
int ans = solution . search( ve , k);
cout << ans << endl;
}
return 0;
}
/*
*/
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-01-21.22
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
class Solution
{
public :
int search( vector < int >& nums , int target)
{
int len = nums . size();
int low1 = 0 , high1 = len - 1 , low2 = 0 , high2 = len - 1;
for( int i = 1; i < len; ++ i)
{
if( nums [ i ] < nums [ i - 1 ])
{
high1 = i - 1 , low2 = i;
break;
}
}
if( high1 == len - 1 && low2 == 0)
return isExist( nums , low1 , high2 , target);
else
{
int idx1 = isExist( nums , low1 , high1 , target);
int idx2 = isExist( nums , low2 , high2 , target);
if( idx1 ==- 1 && idx2 ==- 1)
return - 1;
else
return idx1 !=- 1 ? idx1: idx2;
}
}
int isExist( vector < int >& nums , int low , int high , int target)
{
while( low <= high)
{
int mid =( low + high) >> 1;
if( target < nums [ mid ])
high = mid - 1;
else if( target > nums [ mid ])
low = mid + 1;
else return mid;
}
return - 1;
}
};
int main()
{
Solution solution;
int n , k;
vector < int > ve;
while( cin >>n >> k)
{
int tempVal;
for( int i = 0; i <n; ++ i)
{
cin >> tempVal;
ve . push_back( tempVal);
}
int ans = solution . search( ve , k);
cout << ans << endl;
}
return 0;
}
/*
*/
