Median of Two Sorted Arrays

http://www.lintcode.com/en/problem/median-of-two-sorted-arrays/

Question

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays.

Example

Given A=[1,2,3,4,5,6] and B=[2,3,4,5], the median is 3.5.

Given A=[1,2,3] and B=[4,5], the median is 3.

Analysis

参考九章中的解释

对于一个长度为n的已排序数列a,若n为奇数,中位数为a[n / 2 + 1], 若n为偶数,则中位数(a[n / 2] + a[n / 2 + 1]) / 2; 如果我们可以在两个数列中求出第K小的元素,便可以解决该问题; 不妨设数列A元素个数为n,数列B元素个数为m,各自升序排序,求第k小元素; 取A[k / 2] B[k / 2] 比较; 如果 A[k / 2] > B[k / 2] 那么,所求的元素必然不在B的前k / 2个元素中(证明反证法); 反之,必然不在A的前k / 2个元素中,于是我们可以将A或B数列的前k / 2元素删去,求剩下两个数列的; k - k / 2小元素,于是得到了数据规模变小的同类问题,递归解决; 如果 k / 2 大于某数列个数,所求元素必然不在另一数列的前k / 2个元素中,同上操作就好。

相关: Data Stream Median

Solution

class Solution {
    /**
     * @param A: An integer array.
     * @param B: An integer array.
     * @return: a double whose format is *.5 or *.0
     */
    public double findMedianSortedArrays(int[] A, int[] B) {
        int totalLength = A.length + B.length;
        if (totalLength % 2 == 1) {
            return findKth(A, 0, B, 0, totalLength / 2 + 1);
        }
        return (findKth(A, 0, B, 0, totalLength / 2) + findKth(A, 0, B, 0, totalLength / 2 + 1)) / 2.0;
    }

    public int findKth(int[] A, int A_start, int[] B, int B_start, int k) {
        if (A_start >= A.length) {
            return B[B_start + k - 1];
        }
        if (B_start >= B.length) {
            return A[A_start + k - 1];
        }

        if (k == 1) {
            return Math.min(A[A_start], B[B_start]);
        }


        int A_key = A_start + k / 2 - 1 < A.length
                ? A[A_start + k / 2 - 1]
                : Integer.MAX_VALUE;
        int B_key = B_start + k / 2 - 1 < B.length
                ? B[B_start + k / 2 - 1]
                : Integer.MAX_VALUE;

        if (A_key < B_key) {
            return findKth(A, A_start + k / 2, B, B_start, k - k / 2);
        } else {
            return findKth(A, A_start, B, B_start + k / 2, k - k / 2);
        }

    }
}

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