LeetCode 4. Median of Two Sorted Arrays(java)

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)).

我的思路：把两个array merge以后再找中位数，时间O(M+N), 空间O(n+m). 第二段代码是early stop, 时间和空间都减半，但数量级不变。因为只需要找中位数，所以可以merge到一半就停下来。

代码一：

public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1 == null || nums2 == null) return -1;
int n = nums1.length, m = nums2.length;
int[] nums = new int[m + n];
int index1 = 0, index2 = 0, index = 0;
while (index1 < nums1.length && index2 < nums2.length) {
if (nums1[index1] > nums2[index2]) {
nums[index++] = nums2[index2++];
} else {
nums[index++] = nums1[index1++];
}
}
while (index1 < nums1.length) {
nums[index++] = nums1[index1++];
}
while (index2 < nums2.length) {
nums[index++] = nums2[index2++];
}
if (nums.length % 2 == 1) {
return nums[nums.length / 2];
} else {
double result =  ((double)nums[nums.length / 2 - 1] + (double)nums[nums.length / 2]) / (double)2;
return result;
}
}

代码二：

public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1 == null || nums2 == null) return -1;
int n = nums1.length, m = nums2.length;
int volumn = (m + n) / 2 + 1;
int[] nums = new int[volumn];
int index1 = 0, index2 = 0, index = 0;
while (index < volumn && index1 < nums1.length && index2 < nums2.length) {
if (nums1[index1] > nums2[index2]) {
nums[index++] = nums2[index2++];
} else {
nums[index++] = nums1[index1++];
}
}
while (index < volumn && index1 < nums1.length) {
nums[index++] = nums1[index1++];
}
while (index < volumn && index2 < nums2.length) {
nums[index++] = nums2[index2++];
}
if ((m + n) % 2 == 1) {
return nums[volumn - 1];
} else {
double result =  ((double)nums[volumn - 1] + (double)nums[volumn - 2]) / (double)2;
return result;
}
}