Leetcode "4". Median of the different methods for Sorted Arrays--java

Median of Sorted Arrays

This problem is really a bit difficult, think for a long time, see other people's answer is not to understand all of a sudden. The problem is hard , the original title is as follows: there is the sorted arrays  nums1  AND  nums2  of size m and n respectively. Find The median of the sorted arrays. The overall run time complexity should be O (log (M+n)). For a given two sorted arrays, find the median of the whole two arrays. The time complexity requirement is O (log (m+n)). first, the idea:The requirement for the median of the ontology is adjusted to the number of K when the total of two arrays is sorted. Here is an example to illustrate:
figure I, exampleClass A and Class B students are aend and bend, a day sports teacher said to find them in the middle of the K short classmate, so first will be two classes of students according to the order from small to large to stand two teams. Suppose K is for 13. At this time a class sent out the first x short classmate, B class sent to the lower Y students to compare (note: At this time x+y<=k). if X is 6,y is 7. If the 6th student of A is shorter than the 7th classmate of B, that if the two classes in the overall order, that Class A 6 students must be standing in the Class B 7th, up to a class 6th standing in the Class B 7th, up to the whole row of 12th, how can not turn his platoon first k=13. So you can put the Class A top 6 lost, in the remaining two queues inside to find. But it is no longer to find the first K, but to find the first k-x. Because X did the cardinality was lost. of course, the above example, as long as (x+y) <=k the case, X and y as long as is non-negative, that can be. Just for the sake of efficiency, generally go to x=y= (K/2). When a team is left with little or no people, the value is based on the specific circumstances. Second, Java program:

public class Solution {public double findmediansortedarrays (int[] nums1, int[] nums2) {int m = nums1.length;        int n = nums2.length;                int k = m + N;        if ((k&1) ==1) {return find (nums1, 0, M, nums2, 0, N, k/2+1);    }else return (Find (nums1, 0, M, nums2, 0, N, k/2) +find (nums1, 0, M, nums2, 0, N, k/2+1)/2; }//recursive algorithm, shrinking the range of two arrays, and also changing the value of K relative to the upper starting point of the two search interval public double find (int[] A, int astart, int aend, int[] B, int bstart, I NT Bend,int kth) {//1. Unify the short length placed in the previous item of the find function parameter if (aend>bend) return find (B, bstart, bEnd, a, a            Start, Aend, KTH); 2.            Short length is empty, then the answer is equivalent to finding the median of another array if (aend<=0) return B[bstart + kth-1]; 3.                Recursively to the end point, two arrays of Astart and bstart have reached the median position if (kth==1) return min (A[astart],b[bstart]);                int pa = min (kth/2,aend), PB = Kth-pa; if (A[astart + pa-1]< B[bstart +pb-1]) return find (A, asTart+pa, Aend-pa, B, bstart, bEnd, KTH-PA);                    else return find (A, Astart, Aend, B, bstart + pb, BEND-PB,KTH-PB);    } public int min (int a, int b) {return a>b?b:a; }}

Reference:Leetcode–median of Sorted Arrays Java Median of Sorted Arrays (JAVA)

Leetcode "4". Median of the different methods for Sorted Arrays--java