[leetcode]Median of Two Sorted Arrays

4. Median of Two Sorted Arrays

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

这题大致有两种解法，第一种使用中位数的定义（将有序数组分为同等两部分）然后通过二分法寻找满足中位数的切割。

第二种解法比较通俗易懂，利用另外一个问题：在两个有序数组中找k-th小的数。

Let’s say we have two arrays:

a0,a1,a2,...,am−1

b0,b1,b2,...,bn−1

The array after merging two arrays:

c0,c1,c2,...,cm+n−1

goal: find kth smallest number of merged array, which is ck−1

Theorem: say if ak/2−1<bk/2−1, then numbers ak/2−1 and its left side elements are all smaller than kth smallest number of the merged array:ck−1.

ak/2−1 and its left side elements: a0,a1,...,ak/2−1.

Assume that ck−1=ak/2−1:

In the merged array, there are k−1 elements smaller thanck−1.

In the first array, there are k/2−1 elements smaller than ck−1=ak/2−1.

In the second array, there are at most k/2−1 elements smaller than ck−1=ak/2−1<bk/2−1.

Thus there are at most k−2 elements smaller thanck−1, which is a contradiction.

How to use this theorem?

find(a,b,k)

Compareak/2−1 with bk/2−1,

if ak/2−1<bk/2−1, return find(a+k/2,b,k/2)

else return find(a,b+k/2,k/2)

So the median of two sorted arrays could be solved in O(log(m+n)).

class Solution
{
private:
int k_th(vector<int>::iterator a, int m, vector<int>::iterator b, int n, int k)
{
if (m < n) return k_th(b, n, a, m, k);
if (n == 0) return *(a + k - 1);
if (k == 1) return min(*a, *b);

int j = min(k / 2, n);
int i = k - j;
if (*(a + (i - 1)) > *(b + (j - 1)))
return k_th(a, i, b + j, n - j, k - j);
return k_th(a + i, m - i, b, j, k - i);
}
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
{
int m = nums1.size(), n = nums2.size();
int m1 = k_th(nums1.begin(), m, nums2.begin(), n, (m + n) / 2 + 1);
if ((m + n) % 2 == 0)
return (m1 + k_th(nums1.begin(), m, nums2.begin(), n, (m + n) / 2)) / 2.0;
return m1;
}
};