LeetCode-java实现-T4Median 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)).

注意：这里要注意的是当m+n是偶数是，返回的是中间两个数求和除以2.0的结果，奇数就是返回中间值。

这里采用的是增加一个辅助空间的方式。

package org.algorithm.leetcode;

public class Item4 {
	public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
		int length1 = nums1.length;
		int length2 = nums2.length;
		int medianLength = (length1 + length2)/2;
		
		if(length1 == 0 || nums1 == null)
			if((length1 + length2)%2 == 0)
				return (nums2[medianLength-1]+nums2[medianLength])/2.0;
			else
				return nums2[medianLength];
		else if(length2 == 0 || nums2 == null)
			if((length1 + length2)%2 == 0)
				return (nums1[medianLength-1]+nums1[medianLength])/2.0;
			else
				return nums1[medianLength];
		
		int[] array = new int[length1 + length2];
		int i = 0;
		int j = 0;
		int k = 0;
		while(i < length1 && j < length2) {
			if(nums1[i] <= nums2[j])
				array[k++] = nums1[i++];
			else
				array[k++] = nums2[j++];
		}
		while(i == length1 && j <length2)
			array[k++] = nums2[j++];
		while(j == length2 && i < length1)
			array[k++] = nums1[i++];
		if((length1 + length2)%2 == 0)
			return (array[medianLength-1]+array[medianLength])/2.0;
		else
			return array[medianLength];
    }
	
	public static void main(String[] args) {
		int nums1[] = {};
		int nums2[] = {2,3};
		System.out.println(findMedianSortedArrays(nums1,nums2));
	}
}