[Leetcode-java] Median of Sorted Arrays

Topic:

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)).

Ideas: The main idea is to find the median number of two arrays, check the relevant definition, the calculation of a finite number of data in the median method is: all the same data in the order of size. If the number of data is odd , then the middle data is the median of this group of data, if the number of data is even, then the median of the 2 data is the median of this group of data.

Share the online very concise idea of the algorithm, find two array of the K small element, the core of the method is to turn the original problem into a search for the K decimal point (assuming the two original sequence in ascending order), so the median is actually the first (m+n)/2 small number. So as long as the problem of k decimal is solved, the original problem is solved.

First, assuming that the number of elements of arrays A and b are greater than K/2, we compare a[k/2-1] and b[k/2-1] two elements, which represent the K/2 small element of a and the K/2 small element of B respectively. 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.

Code:

 Public classSolution { Public DoubleFindmediansortedarrays (int[] nums1,int[] nums2) {        intLen1 =nums1.length; intLen2 =nums2.length; intTotal = Len1 +Len2; if(total% 2 = = 1)            returnFindkth (nums1, 0, Len1-1, nums2, 0, len2-1, total/2+1); Else            return(Findkth (nums1, 0, Len1-1, nums2, 0, len2-1, TOTAL/2)                     + findkth (nums1, 0, Len1-1, nums2, 0, len2-1, total/2+1)/2; }         Public DoubleFindkth (int[] A,intAstart,intAend,int[] B,intbstart,intBendintk) {        intm = Aend-astart + 1; intn = bend-bstart + 1; if(M > N)//always keep A for the shortest easy follow-up judgment            returnfindkth (b, bstart, bend, A, Astart, Aend, K); if(M = = 0)//When there is an array that is empty            returnB[bstart + k-1];//Section K Small        if(k = = 1)//Find the smallest            returnmath.min (A[astart], B[bstart]); intda = Math.min (K/2, M);//when the length of M is less than K, the entire a array is required        intdb = K-da; if(A[astart + Da-1] < B[bstart + db-1]){            returnFindkth (A, Astart+da, aend, B, bstart, bend, K-da); }Else if(A[astart + da-1] > B[bstart + db-1]){            returnFindkth (A, Astart, Aend, B, bstart+db, Bend, K-db); }Else{            returnA[astart + da-1]; }    }}

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

Http://www.cnblogs.com/springfor/p/3861890.html

[Leetcode-java] Median of Sorted Arrays