[LeetCode]Median of Two Sorted Arrays

最新推荐文章于 2023-06-12 16:30:30 发布

Jet_yingjia

最新推荐文章于 2023-06-12 16:30:30 发布

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本文介绍了一个高效算法来找到两个已排序数组的中位数，其时间复杂度为O(log(m+n))。该算法通过递归地比较数组的中间元素，缩小搜索范围，最终确定中位数的位置。

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There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

class Solution {
public:
    double findKth(int A[], int m, int B[], int n, int k)
    {
        //always m<=n
        if(m > n)
        {
           return findKth(B, n, A, m, k);
        }
        
        if(m == 0)
        {
            return B[k -1];
        }
        if(k == 1)
        {
            return min(A[0], B[0]);
        }
        //if A[k/2-1] < B[k/2-1], that means all the elements from A[0] to A[k/2-1]
        //(i.e. the k/2 smallest elements in A) are in the range of k smallest elements in the union of A and B. 
        //Or, in the other word, A[k/2 - 1] can never be larger than the k-th smalleset element in the union of A and B.
        int posA = min(m, k/2);
        int posB = k - posA;
        if(A[posA - 1] < B[posB - 1])
        {
            return findKth(A + posA, m - posA, B, n, k - posA);
        }
        else if(A[posA -1] > B[posB - 1])
        {
            return findKth(A, m, B + posB, n - posB, k - posB);
        }
        else
        {
            return A[posA - 1];
        }
    }

    double findMedianSortedArrays(int A[], int m, int B[], int n) {
      if((m+n)&0x01)//odd
      {
          return findKth(A, m, B, n,(m + n)/2 + 1);
      }
      else //even
      {
          return (findKth(A, m, B, n, (m + n)/2) + findKth(A, m, B, n, (m + n)/2 + 1))/2;
      }
        
        
    }
};