设X[0:n - 1 ] 和 Y[0:n - 1 ] 为两个数组，每个数组含有 n 个已经 排好序的数，试设计一个O(log n ) 时间的算法，找出X 和 Y 的 2n个数的中位数

public class Median_of_two_sorted_arrays {//
    public static void main(String args[]){
        //        int []arr1 = {1,4,7,9};
        //        int []arr2 = {2,3,5,6};

        int []arr1 = {1,4,6,8,11};
        int []arr2 = {2,3,5,7,9};

        //        int []arr1 = {1,2,3,4,5};
        //        int []arr2 = {6,7,8,9,10};

        //        int []arr1 = new int[4];
        //        int []arr2 = new int[4];
        //        Scanner scanner = new Scanner(System.in);
        //        for(int i = 0;i < 4;i++){
        //            arr1[i] = scanner.nextInt();
        //        }
        //        for(int i = 0;i < 4;i++){
        //            arr2[i] = scanner.nextInt();
        //        }
        System.out.println("利用分治法求两个有序数组的中位数为：" + findMedianSortedArrays(arr1,arr2));
    }
    public static double findMedianSortedArrays(int []arr1,int []arr2){
        int n = arr1.length;
        int m = arr2.length;
        int L1 = 0,L2 = 0,R1 = 0,R2 = 0,c1,c2,left = 0,right = n - 1;
        c1 = (left + right) / 2;
        while (left <= right){
            c2 = n - c1;
            L1 = (c1 == 0)?arr1[0]:arr1[c1 - 1];//考虑一方数组完全大于另一数组情况
            R1 = (c1 == n - 1)?arr1[n - 1]:arr1[c1];
            L2 = (c2 == 0)?arr2[0]:arr2[c2 - 1];
            R2 = (c2 == n - 1)?arr2[n - 1]:arr2[c2];

            if(c1 == 0){//当arr1完全大于arr2的情况
                return (double)(R2 + L1) /2;
            }
            else if(c1 == n - 1) {//当arr1完全小于arr2的情况
                return (double)(L2 + R1) /2;
            }

            if(L1 > R2){
                c1--;
            }
            else if(L2 > R1){
                c1++;
            }
            else {
                break;
            }
        }
        return ((double)(max(L1,L2)) + min(R1,R2)) / 2;
    }
    public static int max(int a,int b){
        return a > b?a:b;
    }
    public static int min(int a,int b){
        return a < b?a:b;
    }
}