LeetCode: Median of Two Sorted Arrays

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本文介绍了一种在O(log(m+n))的时间复杂度内找到两个已排序数组A和B中位数的方法。通过定义特定的递归函数fink来实现这一目标，并详细展示了如何通过比较和排除部分元素来逐步逼近中位数。

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There are two sorted arrays A and B 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)).

#define INT_MIN 0X80000000
#define INT_MAX 0X7FFFFFFF
class Solution {
public:
    int fink(int A[], int m, int B[], int n, int k)
    {
        if (m <= 0)
            return B[k-1];
        if (n <= 0)
            return A[k-1];
            
        int i = (double)m/(m+n)*k - 1;
    	int j = (k-1) - i;
    
        int Ai_1 = ((i == 0) ? INT_MIN : A[i-1]);
        int Bj_1 = ((j == 0) ? INT_MIN : B[j-1]);
        int Ai   = ((i == m) ? INT_MAX : A[i]);
        int Bj   = ((j == n) ? INT_MAX : B[j]);
  
    	if (Bj_1 <= Ai && Ai <= Bj)
            return Ai;
        else if (Ai_1 <= Bj && Bj <= Ai)
            return Bj;
    
    	//A[i] is too small, get rid of lower part of A and higher part of B
        if (Ai < Bj)
            return fink(A+i+1, m-i-1, B, j, k-i-1);
        
        //B[j] is too small, get rid of higher part of A and lower part of B
        else //if(i > 0 && B[j] < A[i-1]) 
            return fink(A, i, B+j+1, n-j-1, k-j-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 total = m + n;
        if (total % 2 != 0)
            return fink(A, m, B, n, total/2+1);
        else
        {
            double a = fink(A, m, B, n, total/2);
            double b = fink(A, m, B, n, total/2+1);
            return (a + b)/2.0;
        }
    }
};