LeetCode - 33. Search in Rotated Sorted Array

33. Search in Rotated Sorted Array

Problem's Link

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[b]Mean:[/b]

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.

[b]analyse:[/b]

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.

[b]Time complexity: O(N)[/b]

[b]view code[/b]

/**
* -----------------------------------------------------------------
* 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;
}
/*

*/