# Median of Two Sorted Arrays

4. Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:

```
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
```

Example 2:

```
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
```

```
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1 == null || nums2 == null) {
return -1;
}
int m = nums1.length;
int n = nums2.length;
if ((m + n) % 2 != 0) {
return (double)findKth(nums1, nums2, (m + n) / 2, 0, m - 1, 0, n - 1);
} else {
return (findKth(nums1, nums2, (m + n) / 2, 0, m - 1, 0, n - 1)
+ findKth(nums1, nums2, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;
}
}
int findKth(int[] A, int[] B, int k, int aStart, int aEnd, int bStart, int bEnd) {
int aLen = aEnd - aStart + 1;
int bLen = bEnd - bStart + 1;
if (aLen == 0) {
return B[bStart + k];
}
if (bLen == 0) {
return A[aStart + k];
}
if (k == 0) {
return A[aStart] < B[bStart] ? A[aStart] : B[bStart];
}
int aMid = aLen * k / (aLen + bLen);
int bMid = k - aMid - 1;
aMid = aStart + aMid;
bMid = bStart + bMid;
if (A[aMid] > B[bMid]) {
k = k - (bMid - bStart + 1);
aEnd = aMid;
bStart = bMid + 1;
} else {
k = k - (aMid - aStart + 1);
bEnd = bMid;
aStart = aMid + 1;
}
return findKth(A, B, k, aStart, aEnd, bStart, bEnd);
}
}
```