leetcode--Median of Two Sorted Arrays

1.题目描述

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)).

2.解法分析

我本来直接想用二分法求出结果，然后有了下面这个结果，结果发现当m+n为偶数时计算出来的结果不对，因为我最初的算法就是求中位数，而不是题目要求的两个中位数的平均值。

class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
// Start typing your C/C++ solution below
 // DO NOT write int main() function
 unordered_set<int> myHash;
 vector<int>::iterator iter;
 vector<int> result;
 result.assign(2,0);
 if(target%2==0)
 {
     int count=0;
     for(int i=0;i<numbers.size();++i)
     {
         if(numbers[i]==target/2)
         {
             if(count==0){result[0]=i+1;count++;
         }
         else
         {
             result[1]=i+1;return result;
         }
     }
 }
 }
 for(iter=numbers.begin();iter!=numbers.end();++iter)
 {
    myHash.insert(*iter);        }
     for(int i=0;i<numbers.size();++i)
     {
         myHash.erase(numbers[i]);
         if(myHash.count(target-numbers[i])==1)
         {
             result[0]=i+1;break;
         }
         myHash.insert(numbers[i]);
     }
     for(int i=result[0];i<numbers.size();++i)
     {
     if((numbers[result[0]-1]+numbers[i])==target)
     {
          result[1]=i+1;break;
      }
      }
  return result;
 }
 };

 

本来想修修补补，看看能不能够AC，结果弄了好半天烦死了，因此觉得应该会有比较统一的方法。结果搜寻到了一个很统一的方法，这个方法将找两个数组的中位数推广到了找两个数组中第k大的数字，有了这个思路代码就很简单了。

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
        }
    }
};