There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays.
Example
Given A=[1,2,3,4,5,6] and B=[2,3,4,5], the median is 3.5.
Given A=[1,2,3] and B=[4,5], the median is 3.
Challenge
The overall run time complexity should be O(log (m+n)).
LeetCode上的原题，请参见我之前的博客Median of Two Sorted Arrays。
解法一：
class Solution { public: /** * @param A: An integer array. * @param B: An integer array. * @return: a double whose format is *.5 or *.0 */ double findMedianSortedArrays(vector<int> A, vector<int> B) { int n1 = A.size(), n2 = B.size(); if (n1 < n2) return findMedianSortedArrays(B, A); if (n2 == 0) return ((double)A[(n1 - 1) / 2] + (double)A[n1 / 2]) / 2; int left = 0, right = n2 * 2; while (left <= right) { int mid2 = (left + right) / 2; int mid1 = n1 + n2 - mid2; double L1 = mid1 == 0 ? INT_MIN : A[(mid1 - 1) / 2]; double L2 = mid2 == 0 ? INT_MIN : B[(mid2 - 1) / 2]; double R1 = mid1 == n1 * 2 ? INT_MAX : A[mid1 / 2]; double R2 = mid2 == n2 * 2 ? INT_MAX : B[mid2 / 2]; if (L1 > R2) left = mid2 + 1; else if (L2 > R1) right = mid2 - 1; else return (max(L1, L2) + min(R1, R2)) / 2; } return -1; } };
解法二：
class Solution { public: /** * @param A: An integer array. * @param B: An integer array. * @return: a double whose format is *.5 or *.0 */ double findMedianSortedArrays(vector<int> A, vector<int> B) { int m = A.size(), n = B.size(), left = (m + n + 1) / 2, right = (m + n + 2) / 2; return (findKth(A, B, left) + findKth(A, B, right)) / 2.0; } double findKth(vector<int> A, vector<int> B, int k) { int m = A.size(), n = B.size(); if (m > n) return findKth(B, A, k); if (m == 0) return B[k - 1]; if (k == 1) return min(A[0], B[0]); int i = min(m, k / 2), j = min(n, k / 2); if (A[i - 1] > B[j - 1]) { return findKth(A, vector<int>(B.begin() + j, B.end()), k - j); } else { return findKth(vector<int>(A.begin() + i, A.end()), B, k - i); } return -1; } };
本文转自博客园Grandyang的博客，原文链接：两个有序数组的中位数[LintCode] Median of Two Sorted Arrays ，如需转载请自行联系原博主。
