* * Median of Sorted Arrays

There is 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)).

 Public classSolution { Public DoubleFindmediansortedarrays (intA[],intb[]) {        intLen = A.length +b.length; if(len% 2 = = 1) {            returnFindkth (A, 0, B, 0, LEN/2 + 1); }        return(findkth (A,0, B, 0, LEN/2) + findkth (A, 0, B, 0, LEN/2 + 1)        ) /2.0; }    //find kth number of sorted array     Public Static intFindkth (int[] A,intA_start,int[] B,intB_start,intk) {                if(A_start >=a.length) {returnB[b_start + k-1]; }        if(B_start >=b.length) {returnA[a_start + k-1]; }        if(k = = 1) {            returnmath.min (A[a_start], B[b_start]); }                intA_key = A_start + K/2-1 <A.length? A[a_start + K/2-1]: integer.max_value; intB_key = B_start + K/2-1 <B.length? B[b_start + K/2-1]: Integer.max_value;
/* First assume that the number of elements of arrays A and b is greater than K/2, we compare a[k/2-1] and b[k/2-1] Two elements, these two elements represent the first K/2 small element of a and the K/2 small element of B. These two elements compare a total of three cases:>, < and =. If a[k/2-1]<b[k/2-1], this means that the elements of a[0] to a[k/2-1] are in the first K-small element after the merging of a and B. In other words, a[k/2-1] cannot be larger than the K-decimal value after the two array is merged, so we can discard it. */

        if(A_key <B_key) {            returnFindkth (A, A_start + K/2, B, B_start, K-k/2); } Else {            returnFindkth (A, A_start, B, B_start + K/2, K-k/2); }    }}

Reference

http://blog.csdn.net/yutianzuijin/article/details/11499917

* * Median of Sorted Arrays