求两个排好序的数组的中位数 - 二分法

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

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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.

Challenge 

The overall run time complexity should be O(log (m+n)).

这是Princeton算法公开课的课后习题。 资源：http://www.lintcode.com/en/problem/median-of-two-sorted-arrays/

用二分法很好想，找到a[]和b[]的各自中位数比大小，aUb的中位数肯定在这两个数之间，这样就砍掉了一半，实现二分...

然而实施起来的细节tricky到令我发指（见注释）

主要有两点，

第一，假设a比b短，那么a在1,2对应b奇数，偶数，甚至2个，都是不同的corner case，要各自单独小心处理，把每一个case都弄清楚，毕竟奇数个和偶数个的中位数计算方法是不一样的，并且数组太短的话general method可能会越界。

第二，不是每一次都各砍掉一半的！因为这可能改变aUb的奇偶性，从而导致中位数计算方法改变。安全同时比较有效的做法是，a与b每次都砍掉相同的长度，也就是a的一半。

import java.util.Arrays;import java.lang.Math;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 m = A.length;        int n = B.length;        if (m <= n)            return findMedianAB(A, m, B, n);        return findMedianAB(B, n, A, m);    }    private double findMedianAB(int[] A, int m, int[] B, int n) {        assert(m <= n);        if (m == 0) {            return median(B, n);        }        if (m == 1) {            if (n == 1)                return MO2(A[0], B[0]);            if (n % 2 == 1) {                return (MO3(B[n/2 - 1], B[n/2 +1], A[0]) + B[n/2]) / 2.0;            }            return MO3(B[n/2 - 1], B[n/2], A[0]);        }        if (m == 2) {            if (n == 2) {                return MO4(A[0], A[1], B[0], B[1]);            }            if (n % 2 == 1) {                return MO3( B[n/2],                        Math.max(A[0], B[n/2 - 1]),                        Math.min(A[1], B[n/2 + 1]) );            }            return MO4( B[n/2], B[n/2 - 1],                    Math.max( A[0], B[n/2 - 2] ),                    Math.min( A[1], B[n/2 + 1] ) );        }        int midA = (m - 1) / 2;        int midB = (n - 1) / 2;        if (A[midA] <= B[midB]) { // comparing the medians is also ok but a bit slower            return findMedianAB( Arrays.copyOfRange(A, midA, m), m - midA,                    Arrays.copyOfRange(B, 0, n - midA), n - midA );        }        // the most tricky part. reduce the length of BOTH A AND B by midA, rather than by half!        return findMedianAB( Arrays.copyOfRange(A, 0, m - midA), m - midA,                Arrays.copyOfRange(B, midA, n), n - midA );        // try [1,5,6], [2,3,4,7,8] to find that dropping both arrays by half is wrong.        // drop-by-midA approach guarantees two things:        // 1, at least we are still dropping elements safely (only that we drop a bit less elements from B)        // 2, we do not change the being-odd-or-even for the union of sub-A and sub-B from the original one,        // which is important because we need to compute the median of the subset in the same way        // as we do for the original union of A and B (odd: middle; even: average of 2 middles)    }    private double MO2(int a, int b) {        return (a + b) / 2.0;    }    private double MO3(int a, int b, int c) {        return a + b + c                - Math.max(a, Math.max(b, c))                - Math.min(a, Math.min(b, c));    }    private double MO4(int a, int b, int c, int d) {        int max = Math.max( a, Math.max( b, Math.max( c, d ) ) );        int min = Math.min( a, Math.min( b, Math.min( c, d ) ) );        return (a + b + c + d - max - min) / 2.0;    }    private double median(int[] arr, int n) {        if (n == 0)            return Double.NaN; // error        if (n % 2 == 0)            return (arr[n / 2] + arr[n / 2 - 1]) / 2.0;        return arr[n/2];    }}