LeetCodeOJ--002(C++ && Python)

Median of Two Sorted Arrays

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)).

C++版本：

class Solution {

public:
  //快速排序
  void quick_sort(vector<int> &s, int l, int r)
  {
    if (l < r)
    {
      int i = l, j = r, x = s[l];
      while (i < j)
      {
        while(i < j && s[j] >= x) // 从右向左找第一个小于x的数
          j--;  
        if(i < j) 
          s[i++] = s[j];

        while(i < j && s[i] < x) // 从左向右找第一个大于等于x的数
          i++;  
        if(i < j) 
          s[j--] = s[i];
      }
      s[i] = x;
      quick_sort(s, l, i - 1); // 递归调用 
      quick_sort(s, i + 1, r);
    }
  }

public:
  double findMedianSortedArrays(int A[], int m, int B[], int n) 
  {
    if (A == NULL || B == NULL || (m <= 0 && n <= 0))
    {
      return 0.0;
    }

    vector<int> C;
    for (size_t i = 0; i < m + n; i++)
    {
      if (i < m)
      {
        C.push_back(A[i]);
      } 
      else
      {
        C.push_back(B[i-m]);
      }
    }

    quick_sort(C, 0, m+n-1);
    
    double Median = 0.0;

    if ((m+n)&0x01 == 1)
    {
      Median = 1.0*C[(m+n)/2];
    } 
    else
    {
      Median = 1.0*(C[(m+n-1)/2] + C[(m+n)/2])/2;
    }
    return Median;  
  }
};

Python版本：

class Solution:
    # @return a float
    def findMedianSortedArrays(self, A, B):
        L = A + B
        L.sort()
        l = len(L)
        
        Median = 0.0
        
        if l == 0:
            return Median
        else:
            if l*0x01 == 1:
                Median = 1.0*L[l/2]
            else:
                Median = (L[(l-1)/2]+L[l/2])/2.0
            return Median