leetcode Median of Two Sorted Arrays

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)).

 比较容易想到的做法是O(n)，merge两个数组，然后求中值。

public class Solution {
    public double findMedianSortedArrays(int A[], int B[]) {
        // Start typing your Java solution below
        // DO NOT write main() function
        double result = 0;
        int lengthA = A.length;
        int lengthB = B.length;
        
        int[] combinedArray = new int[lengthA + lengthB];
        for(int i = 0; i < lengthA; i++){
            combinedArray[i] = A[i];
        }
        for(int j = 0; j < lengthB; j++){
            combinedArray[lengthA + j] = B[j];
        }
        
        Arrays.sort(combinedArray);
        
        if(((lengthA + lengthB) % 2) !=0){
            result = combinedArray[(lengthA + lengthB) / 2];
           
        } else {
            result = (combinedArray[(lengthA + lengthB) / 2] + combinedArray[(lengthA + lengthB) / 2 - 1]) / 2.0;
            
        }
        return result;
    }
}

 O(log(m+n))第一映像是使用二分搜索

public class Solution {
    public double findMedianSortedArrays(int A[], int B[]) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int aLen = A.length;
        int bLen = B.length;
        if((aLen + bLen) % 2 ==0){
            return (getKthElement(A, 0, aLen - 1, B, 0, bLen - 1, (aLen + bLen) / 2) + getKthElement(A, 0, aLen - 1, B, 0, bLen - 1, (aLen + bLen) / 2 + 1)) / 2.0;
        } else {
            return getKthElement(A, 0, aLen - 1, B, 0, bLen - 1, (aLen + bLen) / 2 + 1);
        }
    }
    
    public int getKthElement(int A[], int aBeg, int aEnd, int B[], int bBeg, int bEnd, int k){
        if(aBeg > aEnd){
            return B[bBeg + (k - 1)];
        }
        if(bBeg > bEnd){
            return A[aBeg + (k - 1)];
        }
        
        int aMid = (aBeg + aEnd) >> 1;
        int bMid = (bBeg + bEnd) >> 1;
        int len = aMid - aBeg + bMid - bBeg + 2;
        
        if(len > k){
            if(A[aMid] < B[bMid]){
                return getKthElement(A, aBeg, aEnd, B, bBeg, bMid - 1, k);
            } else {
                return getKthElement(A, aBeg, aMid - 1, B, bBeg, bEnd, k);
            }
        } else {
            if(A[aMid] < B[bMid]){
                return getKthElement(A, aMid + 1, aEnd, B, bBeg, bEnd, k - (aMid - aBeg + 1));
            } else {
                return getKthElement(A, aBeg, aEnd, B, bMid + 1, bEnd, k - (bMid - bBeg + 1));
            }
        }
    } 
}

(a + b) >> 1 + 1

>> 的优先级比 + 要低，上述等价于(a + b) >> (1 + 1)

ref http://www.cnblogs.com/longdouhzt/archive/2013/03/04/2943572.html