LeetCode: Median of Two Sorted Arrays

这题难度在编程，看了网上答案

class Solution {
public:
    int dfs(int A[], int m, int B[], int n, int k)  
    { 
        if (m <= 0) return B[k-1];  
        if (n <= 0) return A[k-1];  
        if (k <= 1) return min(A[0], B[0]);   
        if (B[n/2] >= A[m/2]  && (m/2+n/2+1 >= k)) return dfs(A, m, B, n/2, k);
        if (B[n/2] >= A[m/2]  && (m/2+n/2+1 < k)) return dfs(A+m/2+1, m-m/2-1, B, n, k-(m/2+1));
        if (B[n/2] < A[m/2]  && (m/2+n/2+1 >= k)) return dfs(A, m/2, B, n, k);
        if (B[n/2] < A[m/2]  && (m/2+n/2+1 < k)) return dfs(A, m, B+n/2+1, n-n/2-1, k-(n/2+1));
    }
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if((n+m)%2)  return dfs(A,m,B,n, (m+n)/2+1);   
        else return (dfs(A, m, B, n, (m+n)/2) + dfs(A, m, B, n, (m+n)/2+1))/2.0;
    }
};

再贴段比较容易懂的代码

class Solution {
public:
    double dfs(int A[], int m, int B[], int n, int k) {
        if (!m) return B[k-1];
        if (!n) return A[k-1];
        int *p, *q;
        if (A[m/2] <= B[n/2]) p = A, q = B;
        else {
            p = B, q = A;
            swap(m, n);
        }
        if (k <= m/2+n/2+1) return dfs(p, m, q, n/2, k);
        else return dfs(p+m/2+1, m-m/2-1, q, n, k-m/2-1);
    }
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int k = m + n;
        if (k % 2) return dfs(A, m, B, n, k/2+1);
        else return (dfs(A, m, B, n, k/2+1)+dfs(A, m, B, n, k/2))/2;
    }
};