Median of Two Sorted Arrays
题目描述:
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)).
找两个有序数组的中位数,这并不难。可是题目把时间复杂度降到O(log(m+n)),难度就上去了。
我尝试解答,想到了把问题转换为求两个有序数组的第k小的数,奈何能力有限,最终铩羽而归。
下面贴出一个Accepted Code:
double findKth(int a[], int m, int b[], int n, int k) { //always assume that m is equal or smaller than n if (m > n) return findKth(b, n, a, m, k); if (m == 0) return b[k - 1]; if (k == 1) return min(a[0], b[0]); //divide k into two parts int pa = min(k / 2, m), pb = k - pa; if (a[pa - 1] < b[pb - 1]) return findKth(a + pa, m - pa, b, n, k - pa); else if (a[pa - 1] > b[pb - 1]) return findKth(a, m, b + pb, n - pb, k - pb); else return a[pa - 1]; } double findMedianSortedArrays(int A[], int m, int B[], int n) { int total = m + n; if (total % 2 == 1) return findKth(A, m, B, n, total / 2 + 1); else return (findKth(A, m, B, n, total / 2) + findKth(A, m, B, n, total / 2 + 1)) / 2; }
如果想了解思路来源,可参考原文链接:http://blog.csdn.net/yutianzuijin/article/details/11499917
ps:
oj论坛上有人给出了复杂度更低的算法,但是思路没有上述方法清晰。