[LeetCode]Median of Two Sorted Arrays

最新推荐文章于 2023-06-12 16:30:30 发布

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本文介绍了一种通过获取两个已排序数组中的第k个数来解决中位数问题的方法，利用分治策略实现算法的简洁性和高效性。详细解释了如何在已排序数组中定位特定位置的元素，进而计算中位数，适用于数组长度为偶数或奇数的情况。

class Solution {
//more detail refer to: http://fisherlei.blogspot.com/2012/12/leetcode-median-of-two-sorted-arrays.html
//using the method of getting the kth number in the two sorted array to solve the median problem
//divide-and-conquer
//very clean and concise
public:
	double findMedianSortedArrays(int A[], int m, int B[], int n) {  
		if((n+m)%2 ==0)  
		{  
			return (GetMedian(A,m,B,n, (m+n)/2) + GetMedian(A,m,B,n, (m+n)/2+1))/2.0;  
		}  
		else  
			return GetMedian(A,m,B,n, (m+n)/2+1);        
	}  
	int GetMedian(int a[], int n, int b[], int m, int k)//get the kth number in the two sorted array  
	{  
		//assert(a && b);   
		if (n <= 0) return b[k-1];  
		if (m <= 0) return a[k-1];  
		if (k <= 1) return min(a[0], b[0]);   
		//a: section1 section2
		//b: section3 section4
		if (b[m/2] >= a[n/2])  
		{  
			if ((n/2 + 1 + m/2) >= k)  
				return GetMedian(a, n, b, m/2, k);//abort section 4  
			else  
				return GetMedian(a + n/2 + 1, n - (n/2 + 1), b, m, k - (n/2 + 1)); //abort section 1 
		}  
		else  
		{  
			if ((m/2 + 1 + n/2) >= k)  
				return GetMedian( a, n/2,b, m, k);//abort section 2
			else  
				return GetMedian( a, n, b + m/2 + 1, m - (m/2 + 1),k - (m/2 + 1));//abort section 3
		}  
	}  
};

second time

class Solution {
public:
    int GetKthNumber(int A[], int m, int B[], int n, int k)
    {
        //terminate case
        if(m <= 0) return B[k];
        if(n <= 0) return A[k];
        if(k <= 0) return min(A[0], B[0]);
        //recursion
        //A:s1,median,s2
        //B:s3,median,s4
        if(A[m/2] > B[n/2])//s2>s3
        {
            if(m/2+n/2+1 >= k+1)
            {
                //A:s1
                //B:s3,median,s4
                GetKthNumber(A, m/2, B, n, k);
            }
            else
            {
                //A:s1,median,s2
                //B:s4
                GetKthNumber(A, m, B+(n/2+1), n-(n/2+1), k-(n/2+1));
            }
        }
        else//A[m/2] < B[n/2]
        {
            if(m/2+n/2+1 >= k+1)
                GetKthNumber(A, m, B, n/2, k);
            else
                GetKthNumber(A+(m/2+1), m-(m/2+1), B, n, k-(m/2+1));
        }
    }
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if((m+n)%2 == 0)//even
            return (double)(GetKthNumber(A, m, B, n, (m+n)/2-1)+GetKthNumber(A, m, B, n, (m+n)/2))/2.0;
        else 
            return (double)GetKthNumber(A, m, B, n, (m+n)/2);
    }
};