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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.

//vs2012测试代码

#include<iostream>

using namespace std;

#define N 8

//classified discussion
//1. Based on the property of rotated array, there may or may not have one sorted sequence 
//when one sequence is divided into two parts  
//2. make decision under all these cases
class Solution {
public:
    bool search(int A[], int n, int target) 
	{
        if( n==0 )
			return false;
        int mid = 0;
		int left=0, right=n-1;
		while( left <= right)
		{
			mid = left + (right-left)/2 ;
			if(A[mid] == target)
				return true;
			if( A[left] < A[mid] ) //left side sorted, including mid, it is sorted in [l,mid]
			{
				if( A[left] <= target && target < A[mid] )
					right = mid - 1;
				else
					left = mid + 1;
			}
			else if ( A[mid] < A[right] ) //right side sorted, [mid,r] must be sorted, if [l,mid] not sorted
			{
				if ( A[mid] < target && target <= A[right] )
					left = mid + 1;
				else
					right = mid - 1;
			}
			else if ( A[left] == A[mid] ) //A[m] is not the target, so remove en element equal to A[m] is safe
				left++;
			else if ( A[right] == A[mid] ) //ditto
				right--;
		}
		return false;
    }
};

int main()
{
	int A[N];
	int a , target;
	for(int i=0; i<N; i++)
	{
		cin>>a;
		A[i] = a;
	}
	cin>>target;
	Solution lin;
	cout<<lin.search( A,N,target)<<endl;

	return 0;
}

//方法一：自测Accepted
//classified discussion
//1. Based on the property of rotated array, there may or may not have one sorted sequence 
//when one sequence is divided into two parts  
//2. make decision under all these cases
class Solution {
public:
    bool search(int A[], int n, int target) 
	{
        if( n==0 )
			return false;
        int mid = 0;
		int left=0, right=n-1;
		while( left <= right)
		{
			mid = left + (right-left)/2 ;
			if(A[mid] == target)
				return true;
			if( A[left] < A[mid] ) //left side sorted, including mid, it is sorted in [l,mid]
			{
				if( A[left] <= target && target < A[mid] )
					right = mid - 1;
				else
					left = mid + 1;
			}
			else if ( A[mid] < A[right] ) //right side sorted, [mid,r] must be sorted, if [l,mid] not sorted
			{
				if ( A[mid] < target && target <= A[right] )
					left = mid + 1;
				else
					right = mid - 1;
			}
			else if ( A[left] == A[mid] ) //A[m] is not the target, so remove en element equal to A[m] is safe
				left++;
			else if ( A[right] == A[mid] ) //ditto
				right--;
		}
		return false;
    }
};

leetcode_81_Search in Rotated Sorted Array II

标签：c++   leetcode   binary search   array   

原文地址：http://blog.csdn.net/keyyuanxin/article/details/43638543