[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

這題大致有兩種解法，第一種使用中位數的定義（將有序數組分爲同等兩部分）然後通過二分法尋找滿足中位數的切割。
第二種解法比較通俗易懂，利用另外一個問題：在兩個有序數組中找k-th小的數。
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;
    }
};