4. Median of Two Sorted Arrays
Total Accepted: 99662 Total Submissions: 523759
Difficulty: Hard
There are two sorted arrays nums1 and nums2 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)).
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
方案0：合并两个数组为一个数组，排序，取第k个
class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
{
// Start typing your C/C++ solution below
// DO NOT write int main() function
int m = nums1.size();
int n = nums2.size();
vector<int> v;
v.insert(v.end(), nums1.begin(), nums1.end());
v.insert(v.end(), nums2.begin(), nums2.end());
sort(v.begin(),v.end());
double median=(double) ((n+m)%2? v[(n+m)/2]:(v[(n+m-1)/2]+v[(n+m)/2])/2.0);
return median;
}
};方案1：假设两个数组总共有n个元素，用merge sort的思路排序，排序好的数组取出下标为k-1的元素就是我们需要的答案。
这个方法比较容易想到，但是有没有更好的方法呢？
方案2：可以用一个计数器，记录当前已经找到第m大的元素。同时我们使用两个指针pA和pB，分别指向A和B数组的第一个元素。使用类似于merge sort的原理，如果数组A当前元素小，那么pA++，同时m++。如果数组B当前元素小，那么pB++，同时m++。最终当m等于k的时候，就得到了我们的答案——O(k)时间，O(1)空间。
但是，当k很接近于n的时候，这个方法还是很费时间的。当然，我们可以判断一下，如果k比n/2大的话，我们可以从最大的元素开始找。但是如果我们要找所有元素的中位数呢？时间还是O(n/2)=O(n)的。有没有更好的方案呢？
我们可以考虑从k入手。如果我们每次都能够剔除一个一定在第k大元素之前的元素，那么我们需要进行k次。但是如果每次我们都剔除一半呢？所以用这种类似于二分的思想，我们可以这样考虑：
Assume that the number of elements in A and B are both larger than k/2, and if we compare the k/2-th smallest element in A(i.e. A[k/2-1]) and the k-th smallest element in B(i.e. B[k/2 - 1]), there are three results:
(Becasue k can be odd or even number, so we assume k is even number here for simplicy. The following is also true when k is an odd number.)
A[k/2-1] = B[k/2-1]
A[k/2-1] > B[k/2-1]
A[k/2-1] < B[k/2-1]
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.
Why?
We can use a proof by contradiction. Since A[k/2 - 1] is larger than the k-th smallest element in the union of A and B, then we assume it is the (k+1)-th smallest one. Since it is smaller than B[k/2 - 1], then B[k/2 - 1] should be at least the (k+2)-th smallest one. So there are at most (k/2-1) elements smaller than A[k/2-1] in A, and at most (k/2 - 1) elements smaller than A[k/2-1] in B.So the total number is k/2+k/2-2, which, no matter when k is odd or even, is surly smaller than k(since A[k/2-1] is the (k+1)-th smallest element). So A[k/2-1] can never larger than the k-th smallest element in the union of A and B if A[k/2-1]<B[k/2-1];
Since there is such an important conclusion, we can safely drop the first k/2 element in A, which are definitaly smaller than k-th element in the union of A and B. This is also true for the A[k/2-1] > B[k/2-1] condition, which we should drop the elements in B.
When A[k/2-1] = B[k/2-1], then we have found the k-th smallest element, that is the equal element, we can call it m. There are each (k/2-1) numbers smaller than m in A and B, so m must be the k-th smallest number. So we can call a function recursively, when A[k/2-1] < B[k/2-1], we drop the elements in A, else we drop the elements in B.
We should also consider the edge case, that is, when should we stop?
1. When A or B is empty, we return B[k-1]( or A[k-1]), respectively;
2. When k is 1(when A and B are both not empty), we return the smaller one of A[0] and B[0]
3. When A[k/2-1] = B[k/2-1], we should return one of them
In the code, we check if m is larger than n to garentee that the we always know the smaller array, for coding simplicy.
中文翻译：
该方法的核心是将原问题转变成一个寻找第k小数的问题（假设两个原序列升序排列），这样中位数实际上是第(m+n)/2小的数。所以只要解决了第k小数的问题，原问题也得以解决。
首先假设数组A和B的元素个数都大于k/2，我们比较A[k/2-1]和B[k/2-1]两个元素，这两个元素分别表示A的第k/2小的元素和B的第k/2小的元素。这两个元素比较共有三种情况：>、<和=。如果A[k/2-1]<B[k/2-1]，这表示A[0]到A[k/2-1]的元素都在A和B合并之后的前k小的元素中。换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
证明也很简单，可以采用反证法。假设A[k/2-1]大于合并之后的第k小值，我们不妨假定其为第（k+1）小值。由于A[k/2-1]小于B[k/2-1]，所以B[k/2-1]至少是第（k+2）小值。但实际上，在A中至多存在k/2-1个元素小于A[k/2-1]，B中也至多存在k/2-1个元素小于A[k/2-1]，所以小于A[k/2-1]的元素个数至多有k/2+ k/2-2，小于k，这与A[k/2-1]是第（k+1）的数矛盾。
当A[k/2-1]>B[k/2-1]时存在类似的结论。
当A[k/2-1]=B[k/2-1]时，我们已经找到了第k小的数，也即这个相等的元素，我们将其记为m。由于在A和B中分别有k/2-1个元素小于m，所以m即是第k小的数。(这里可能有人会有疑问，如果k为奇数，则m不是中位数。这里是进行了理想化考虑，在实际代码中略有不同，是先求k/2，然后利用k-k/2获得另一个数。)
通过上面的分析，我们即可以采用递归的方式实现寻找第k小的数。此外我们还需要考虑几个边界条件：
- 如果A或者B为空，则直接返回B[k-1]或者A[k-1]；
- 如果k为1，我们只需要返回A[0]和B[0]中的较小值；
- 如果A[k/2-1]=B[k/2-1]，返回其中一个；
// leetcode4.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#define min(x,y) (x>y?y:x)
#define max(x,y) (x>y?x:y)
double findKth(int a[],int m,int b[],int n,int k)
{
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]);
//divide k into two parts;
int pa = min(k/2,m),pb = k - pa;
if (a[pa -1]<b[pb - 1])
return findKth(a +pa,m-pa,b,n,k-pa);
else if(a[pa -1]>a[pb-1])
return findKth(a,m,b+pb,n-pb,k-pb);
else
return a[pa -1];
}
double findMedianSortedArrays(int A[],int m,int B[],int n)
{
int total = m +n;
if (total&0x1)
return findKth(A,m,B,n,total/2+1);
else
return (findKth(A,m,B,n,total/2)+findKth(A,m,B,n,total/2+1))/2;
}
int _tmain(int argc, _TCHAR* argv[])
{
int a[]={1,2,3};
int b[]={555,666,999};
int result = findMedianSortedArrays(a,3,b,3);
return 0;
}
python解决方案：基本上和c++比较类似
def findMedianSortedArrays(self, A, B):
l = len(A) + len(B)
if l % 2 == 1:
return self.kth(A, B, l // 2)
else:
return (self.kth(A, B, l // 2) + self.kth(A, B, l // 2 - 1)) / 2.defkth(self, a, b, k):ifnot a:
return b[k]
ifnot b:
return a[k]
ia, ib = len(a) // 2 , len(b) // 2
ma, mb = a[ia], b[ib]
# when k is bigger than the sum of a and b's median indices if ia + ib < k:
# if a's median is bigger than b's, b's first half doesn't include kif ma > mb:
return self.kth(a, b[ib + 1:], k - ib - 1)
else:
return self.kth(a[ia + 1:], b, k - ia - 1)
# when k is smaller than the sum of a and b's indiceselse:
# if a's median is bigger than b's, a's second half doesn't include kif ma > mb:
return self.kth(a[:ia], b, k)
else:
return self.kth(a, b[:ib], k)参考文献：
http://blog.csdn.net/zxzxy1988/article/details/8587244
http://blog.csdn.net/yutianzuijin/article/details/11499917
网上看到了一张leetcode 的难度和考试频率分析表，转过来给大家看看，出现频率为5的题目还是背诵并默写吧，哈哈！
