【leetcode刷题笔记】Median of Two Sorted Arrays

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)).

我们实现一个二分在有序两个数组中找第k小的数的函数，然后在主函数中，根据合并后数组元素个数的奇偶性调用这个函数。

首先来看这个在两个有序数组中找第k小的数的函数 private double findKth(int a[],int b[],int k,int a_start,int b_start){ 。它的原理如下图所示：

即：

if(a_mid < b_mid)
    return findKth(a, b, k-k/2, a_start+k/2, b_start);
else
    return findKth(a, b, k-k/2, a_start, b_start+k/2);

再来看求解函数 public double findMedianSortedArrays(int A[], int B[]) ，当A和B合并后元素个数为奇数的时候，我们直接调用 findKth(A, B, (B.length+A.length)/2+1, 0, 0) 找到第(A.length+B.length)/2的数就是中位数了。而当A和B的合并后元素个数为偶数的时候，我们要调用两次findKth分别找到第(A.length+B.length)/2和第(A.length+B.length)/2-1的数，然后求它们的平均数，即 (findKth(A, B, (A.length+B.length)/2, 0, 0) + findKth(A, B, (A.length+B.length)/2+1, 0, 0)) / 2.0 。

代码如下：

 1 public class Solution {
 2     private double findKth(int a[],int b[],int k,int a_start,int b_start){
 3         //if a is empty
 4         if(a_start >= a.length)
 5             return b[b_start+k-1];
 6         if(b_start >= b.length)
 7             return a[a_start+k-1];
 8
 9         if(k == 1)
10             return Math.min(a[a_start], b[b_start]);
11
12         int a_mid = a_start + k/2 -1 < a.length?a[a_start+k/2-1]:Integer.MAX_VALUE;
13         int b_mid = b_start + k/2 -1 < b.length?b[b_start+k/2-1]:Integer.MAX_VALUE;
14
15         if(a_mid < b_mid){
16             return findKth(a, b, k-k/2, a_start+k/2, b_start);
17         }
18         else
19             return findKth(a, b, k-k/2, a_start, b_start+k/2);
20
21     }
22     public double findMedianSortedArrays(int A[], int B[]) {
23         int len = A.length + B.length;
24         if(len % 2 == 0)
25             return (findKth(A, B, len/2, 0, 0) + findKth(A, B, len/2+1, 0, 0)) / 2.0;
26         else
27             return findKth(A, B, len/2+1, 0, 0);
28     }
29 }