public static void main(String[] args) {

int[] A = {1,3,4,7};

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

double res = findMedianSortedArrays(A, B);

System.out.println(res);

}

private static double findMedianSortedArrays(int[] nums1, int[] nums2) {

int n = nums1.length;

int m = nums2.length;

int left = (n + m + 1) / 2;

int right = (n + m + 2) / 2;

//一个小技巧：将偶数和奇数的情况合并，如果是奇数，会求两次同样的 k 。

return (getKth(nums1, 0, n - 1, nums2, 0, m - 1, left) + getKth(nums1, 0, n - 1, nums2, 0, m - 1, right)) * 0.5;

}

private static int getKth(int[] nums1, int start1, int end1, int[] nums2, int start2, int end2, int k) {

int len1 = end1 - start1 + 1;

int len2 = end2 - start2 + 1;

//让 len1 的长度小于 len2，这样就能保证如果有数组空了，一定是 len1

if (len1 > len2) return getKth(nums2, start2, end2, nums1, start1, end1, k);

if (len1 == 0) return nums2[start2 + k - 1];

if (k == 1) return Math.min(nums1[start1], nums2[start2]);

int i = start1 + Math.min(len1, k / 2) - 1;

int j = start2 + Math.min(len2, k / 2) - 1;

if (nums1[i] > nums2[j]) {

return getKth(nums1, start1, end1, nums2, j + 1, end2, k - (j - start2 + 1));

}

else {

return getKth(nums1, i + 1, end1, nums2, start2, end2, k - (i - start1 + 1));

}

}