首页 > 代码库 > 【leetcode】Median of Two Sorted Arrays(hard)★!!
【leetcode】Median of Two Sorted Arrays(hard)★!!
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)).
思路:
难,知道用分治算法,却不知道怎么用。只好看答案。
基本的思路是如果中位数是第K个数,A[i]如果是中位数,那么A[i]已经大于了i个数,还应大于K - i - 1个数 与B[K-i-2]对比。但是如果中位数不在A中我脑子就晕晕的。下面是大神代码,我还是没有看懂。
class Solution {public: double findMedianSortedArrays(int A[], int m, int B[], int n) { // the following call is to make sure len(A) <= len(B). // yes, it calls itself, but at most once, shouldn‘t be // consider a recursive solution if (m > n) return findMedianSortedArrays(B, n, A, m); double ans = 0; // now, do binary search int k = (n + m - 1) / 2; int l = 0, r = min(k, m); // r is n, NOT n-1, this is important!! while (l < r) { int midA = (l + r) / 2; int midB = k - midA; if (A[midA] < B[midB]) l = midA + 1; else r = midA; } // after binary search, we almost get the median because it must be between // these 4 numbers: A[l-1], A[l], B[k-l], and B[k-l+1] // if (n+m) is odd, the median is the larger one between A[l-1] and B[k-l]. // and there are some corner cases we need to take care of. int a = max(l > 0 ? A[l - 1] : -(1<<30), k - l >= 0 ? B[k - l] : -(1<<30)); if (((n + m) & 1) == 1) return (double) a; // if (n+m) is even, the median can be calculated by // median = (max(A[l-1], B[k-l]) + min(A[l], B[k-l+1]) / 2.0 // also, there are some corner cases to take care of. int b = min(l < m ? A[l] : (1<<30), k - l + 1 < n ? B[k - l + 1] : (1<<30)); return (a + b) / 2.0; }};
【leetcode】Median of Two Sorted Arrays(hard)★!!
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。