LeetCode 4 Median of Two Sorted Arrays

LeetCode 4 Median of Two Sorted Arrays

翻譯

有兩個給定的排好序的數組nums1和nums2，其大小分別為m和n。
找出這兩個已排序數組的中位數。
總運行時間的復雜度應該是O(log(m+n))。

原文

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)).

C#

public class Solution {
    public double FindMedianSortedArrays(int[] nums1, int[] nums2) {
        int len1=nums1.Length;
        int len2=nums2.Length;
        bool isEven=(nums1.Length+nums2.Length)%2==0;

        int left=(len1+len2+1)/2;
        int right=(len1+len2+2)/2;

        if (isEven)
        {
            var leftValue = findKth(nums1, 0, len1 - 1, nums2, 0, len2 - 1, left);
            var rightValue = findKth(nums1, 0, len1 - 1, nums2, 0, len2 - 1, right);
            return (leftValue + rightValue) / 2.0;
        }
        else
        {
            return findKth(nums1, 0, len1 - 1, nums2, 0, len2 - 1, right);
        }
    }
    public double findKth(int[] A,int lowA,int highA,int[] B,int lowB,int highB,int k)
    {
        if(lowA>highA)
        {
            return B[lowB+k-1];
        }
        if(lowB>highB)
        {
            return A[lowA+k-1];
        }
        int midA=(lowA+highA)/2;
        int midB=(lowB+highB)/2;
        if (A[midA] <= B[midB])
        {
            return k <= midA - lowA + midB - lowB + 1 ?
                this.findKth(A, lowA, highA, B, lowB, midB - 1, k) :
                this.findKth(A, midA + 1, highA, B, lowB, highB, k - (midA - lowA + 1));
        }
        else
        {
            return k <= midA - lowA + midB - lowB + 1 ?
                this.findKth(A, lowA, midA - 1, B, lowB, highB, k) : 
                this.findKth(A, lowA, highA, B, midB + 1, highB, k - (midB - lowB + 1));
        }
    }
}