Find the Kth element from two sorted arrays

这道题的关键是如何建立两个数组的之间的关系。插入排序法和双指针法可以达到线性时间复杂度。这里记录一个更加巧妙的思路，利用二分法达到O(logm + logn)的时间复杂度。

原文地址 https://articles.leetcode.com/find-k-th-smallest-element-in-union-of/

import java.util.Arrays;

public class FindKthElement {
    public int findKthElement(int[] arr1, int[] arr2, int k) {
        // k>0, assert arr1 != null, arr2 != null, k <= arr1.length + arr2.lengt
        return findKth(arr1, arr2, arr1.length, arr2.length, k);
    }

    private int findKth(int[] arr1, int[] arr2, int m, int n, int k) {
        int i = k * m / (m + n);

        // important relation of two arrays
        int j = k - 1 - i;

        int Ai_1 = (i == 0) ? Integer.MIN_VALUE : arr1[i - 1];
        int Ai = (i == m) ? Integer.MAX_VALUE : arr1[i];
        int Bj_1 = (j == 0) ? Integer.MIN_VALUE : arr2[j - 1];
        int Bj = (j == n) ? Integer.MAX_VALUE : arr2[j];

        if (Ai > Bj_1 && Ai < Bj) {
            return Ai;
        }
        if (Bj > Ai_1 && Bj < Ai) {
            return Bj;
        }

        // assert Ai < Bj && Ai < Bj_1 ||  Bj < Ai && Bj < Ai_1
        if (Ai < Bj) {
            return findKth(Arrays.copyOfRange(arr1, i + 1, m), Arrays.copyOfRange(arr2, 0, j), m - i - 1, j, k - i - 1);
        } else {
            return findKth(Arrays.copyOfRange(arr1, 0, j), Arrays.copyOfRange(arr2, j + 1, n), i, n - j - 1, k - j - 1);
        }
    }

    public static void main(String[] args) {
        FindKthElement findKthElement = new FindKthElement();
        int[] arra = {1, 3, 5, 7, 9};
        int[] arrb = {2, 4, 6, 8, 10};
        System.out.println(findKthElement.findKthElement(arra, arrb, 6));
    }
}