Leetcode - 4. Median of Two Sorted Arrays

タイトル
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)).
Simple solution: O(m+n)

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    int m = nums1.size();
    int n = nums2.size();
    int* a = new int[m+n];
    int p1=0;
    int p2=0;
    int end = (m+n)/2+1;
    int count=0;
    int flag= 0;
    double res=0.0;

    if((m+n)%2==0){
        flag = 1;
    }else{
        flag = 0;
    }
    while(p1<m && p2<n){
        if(nums1[p1]<=nums2[p2]){
            a[count++]=nums1[p1++];
        }else{
            a[count++]=nums2[p2++];
        }

        if(count==end){
            if(flag){
                res = (double(a[count-1])+double(a[count-2]))/2;
            }else{
                res= double(a[count-1]);

            }
            return res;
        }
    }
    if(p1==m){
        while(p2<n){
            a[count++]=nums2[p2++];
        }
    }

    if(p2==n){
        while(p1<m){
            a[count++]=nums1[p1++];
        }
    }

    if(flag){
        res = (double(a[end-1])+double(a[end-2]))/2;
    }else{
        res = double(a[end-1]);
    }
    return res;
}
};

二分法:O(log(m+n))
Leetcodeより抜粋

class Solution {
public:
    int getkth(int s[], int m, int l[], int n, int k){
        // let m <= n
        if (m > n) 
            return getkth(l, n, s, m, k);
        if (m == 0)
            return l[k - 1];
        if (k == 1)
            return min(s[0], l[0]);

        int i = min(m, k / 2), j = min(n, k / 2);
        if (s[i - 1] > l[j - 1])
            return getkth(s, m, l + j, n - j, k - j);
        else
            return getkth(s + i, m - i, l, n, k - i);
        return 0;
    }

    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        //Note:  m+n   ，l==r; m+n   ， l+1==r.
        int l = (m + n + 1) >> 1;
        int r = (m + n + 2) >> 1;
        return (getkth(A, m ,B, n, l) + getkth(A, m, B, n, r)) / 2.0;
    }
};