给定两个有序数组a1，a2，长度分别是m，n。寻找这两个数组的中间点。例如
a1 = [1, 2]
a2 = [3，4]
中点是2.5
算法思路来自https://discuss.leetcode.com/topic/16797/very-concise-o-log-min-m-n-iterative-solution-with-detailed-explanation
以下代码是kotlin版本实现
/**
* Created by thinkreed on 2017/6/23.
*/
fun findMedianOfTwoArray(a1: Array<Int>, a2: Array<Int>): Double {
//保证a2比a1短
if (a1.size < a2.size) return findMedianOfTwoArray(a2, a1)
//a2为空，返回a1的中间点
if (a2.isEmpty()) return (a1[(a1.size - 1) / 2].toDouble() + a1[a1.size / 2].toDouble()) / 2
var low = 0
var high = 2 * a2.size
//二分查找均分点
while (low <= high) {
//以中点切分
val mid2 = (low + high) / 2
//通过mid2推断mid1,若mid1和mid2是两数组的两个切分点，则满足mid1+mid2 = a1.size + a2.size
val mid1 = a1.size + a2.size - mid2
//L1，R1分别表示数组a1的切分点C1的左相邻元素，右相邻元素
//L2,R2分别表示数组a2的切分点C2的左相邻元素，右相邻元素
val L1 = if (mid1 == 0) Int.MIN_VALUE else a1[(mid1 - 1) / 2]
val L2 = if (mid2 == 0) Int.MIN_VALUE else a2[(mid2 - 1) / 2]
val R1 = if (mid1 == 2 * a2.size) Int.MAX_VALUE else a1[mid1 / 2]
val R2 = if (mid2 == 2 * a2.size) Int.MAX_VALUE else a2[mid2 / 2]
//a1的左边部分太大了，将切分点C1左移，C2相应得右移
if (L1 > R2) low = mid2 + 1
//a2的左边部分太大了,将切分点C2左移，C1相应右移
else if (L2 > R1) high = mid2 - 1
//满足条件L1<=R2 && L2<=R1
else return (maxOf(L1, L2).toDouble() + minOf(R1, R2).toDouble()) / 2
}
return -1.0
}
