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Wednesday, December 26, 2012

[LeetCode] Median of Two Sorted Arrays 解题报告


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)).
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[解题思路]
O(n)的解法比较直观,直接merge两个数组,然后求中间值。而对于O(log(m+n))显然是用二分搜索了, 相当于“Kth element in 2 sorted array”的变形。如果(m+n)为奇数,那么找到“(m+n)/2+1 th element in 2 sorted array”即可。如果(m+n)为偶数,需要找到(m+n)/2 th 及(m+n)/2+1 th,然后求平均。

而对于“Kth element in 2 sorted array”, 如下图,两个中位数 A[m/2] 和 B[n/2], 可以将数组划分为四个部分。而丢弃哪一个部分取决于两个条件:1, (m/2 + n/2)?k;2,A[m/2] ? B[n/2];


如果 (m/2 + n/2) > k,那么意味着,当前中位数取高了,正确的中位数要么在 Section 1或者Section3中。如果A[m/2] > B[n/2], 意味着中位数肯定不可能在Section 2里面,那么新的搜索可以丢弃这个区间段了。同理可以推断出余下三种情况,如下所示:

If (m/2+n/2+1) > k && am/2 > bn/2 , drop Section 2
If (m/2+n/2+1) > k && am/2 < bn/2 , drop Section 4
If (m/2+n/2+1) < k && am/2 > bn/2 ,  drop Section 3
If (m/2+n/2+1) < k && am/2 < bn/2drop Section 1


简单的说,就是或者丢弃最大中位数的右区间,或者丢弃最小中位数的左区间。

[Code]

1:    double findMedianSortedArrays(int A[], int m, int B[], int n) {  
2:      if((n+m)%2 ==0)  
3:      {  
4:        return (GetMedian(A,m,B,n, (m+n)/2) + GetMedian(A,m,B,n, (m+n)/2+1))/2.0;  
5:      }  
6:      else  
7:        return GetMedian(A,m,B,n, (m+n)/2+1);        
8:    }  
9:       int GetMedian(int a[], int n, int b[], int m, int k)  
10:       {  
11:            assert(a && b);   
12:            if (n <= 0) return b[k-1];  
13:            if (m <= 0) return a[k-1];  
14:            if (k <= 1) return min(a[0], b[0]);   
15:            if (b[m/2] >= a[n/2])  
16:            {  
17:                 if ((n/2 + 1 + m/2) >= k)  
18:                      return GetMedian(a, n, b, m/2, k);  
19:                 else  
20:                      return GetMedian(a + n/2 + 1, n - (n/2 + 1), b, m, k - (n/2 + 1));  
21:            }  
22:            else  
23:            {  
24:                 if ((m/2 + 1 + n/2) >= k)  
25:                      return GetMedian( a, n/2,b, m, k);  
26:                 else  
27:                      return GetMedian( a, n, b + m/2 + 1, m - (m/2 + 1),k - (m/2 + 1));  
28:            }  
29:       }  

[注意]
递归的终止条件如n==0, m ==0及k==1的处理


[Rank]
算法:***
实现:****






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