[Leetcode] Median of Two Sorted Arrays (Java)

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There are two sorted arrays A and B 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)).

一共有m+n个数，求中位数，分奇偶两种情况：

1）奇数时，中位数为第(m+n)/2+1位；

2）偶数时，中位数为第(m+n)/2位与第(m+n)/2+1位的平均数。

接下来的问题就是求第k位的数的大小：

二分法分两个部分，每个部分尽量达到k/2大小，若不够，则取最大max，与另一个数组的k-max位比较，得出结果。

public class MedianofTwoSortedArrays {private double find(int A[],int B[],int aLow,int aHigh,int bLow,int bHigh,int k){if(aHigh-aLow < bHigh-bLow) return find(B, A, bLow, bHigh, aLow, aHigh, k);if(bHigh == bLow-1) return A[aLow+k-1];if(k == 1)return A[aLow]>B[bLow]?B[bLow]:A[aLow];int bb = Math.min(k/2, bHigh-bLow+1);int aa = k - bb;if(B[bb+bLow-1] < A[aLow+aa-1])return find(A, B, aLow, aHigh, bLow+bb,bHigh, k-bb);else if(B[bb+bLow-1] > A[aLow+aa-1])return find(A, B, aLow+aa, aHigh, bLow, bHigh, k-aa);elsereturn A[aLow+aa-1];}public double findMedianSortedArrays(int A[], int B[]) {int med;boolean flag = true;if((A.length+B.length)%2==1) {med = (A.length+B.length)/2 +1;}else {med = (A.length+B.length)/2 ;flag = false;}if(flag) {if(A.length == 0 && B.length!=0)return B[med-1];else if (A.length!=0 && B.length == 0)return A[med-1];return find(A, B, 0, A.length-1, 0, B.length-1, med);}else {if(A.length == 0 && B.length!=0)return ((double)B[med]+B[med-1])/2;else if (A.length!=0 && B.length == 0)return ((double)A[med]+A[med-1])/2;return (find(A, B, 0, A.length-1, 0, B.length-1, med)+find(A, B, 0, A.length-1, 0, B.length-1, med+1))/2;}}public static void main(String[] args) {int[] a = {1,3,5,7,9,11};int[] b = {2,4,6,8,10,12,14,16};double result = new MedianofTwoSortedArrays().findMedianSortedArrays(a, b);System.out.println(result);}}

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