Java for Leetcode 004 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)).

Problem Solving Ideas:

Due to the requirement of time complexity O (log (m+n)) It is almost certainly the idea of recursion and division.

Since there was an algorithm for finding two decimal places in the array, the time complexity is O (log (m+n)), so the direct call can be

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

Java Reference Code:

 Public classSolution { Public Static DoubleFindkth (int[] nums1,intIndex1,int[] nums2,intIndex2,intk) {if(Nums1.length-index1 > Nums2.length-index2)returnfindkth (NUMS2, Index2, Nums1, index1, K); if(Nums1.length-index1 = = 0)            returnNums2[index2 + k-1]; if(k = = 1)            returnmath.min (Nums1[index1], nums2[index2]); intP1 = Math.min (K/2, nums1.length-index1), p2 = k-P1; if(Nums1[index1 + p1-1] < Nums2[index2 + p2-1])            returnFindkth (nums1, index1 + p1, Nums2, Index2, K-p1); Else if(Nums1[index1 + p1-1] > nums2[index2 + p2-1])            returnFindkth (Nums1, index1, NUMS2, Index2 + p2, K-p2); Else            returnNums1[index1 + p1-1]; }    Static  Public DoubleFindmediansortedarrays (int[] nums1,int[] nums2) {        if((nums1.length + nums2.length)% 2! = 0)            returnFindkth (nums1, 0, NUMS2, 0, (Nums1.length+ nums2.length)/2 + 1); Else            returnFindkth (nums1, 0, NUMS2, 0, (Nums1.length+ nums2.length)/2)                    /2 + findkth (nums1, 0, NUMS2, 0, (Nums1.length+ nums2.length)/2 + 1)/2; }}

Java for Leetcode 004 Median of Sorted Arrays