There are two sorted arrays nums1 and nums2 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)).
Example 1:
nums1 = [1, 3] nums2 = [2] The median is 2.0
Example 2:
nums1 = [1, 2] nums2 = [3, 4] The median is (2 + 3)/2 = 2.5
这个问题可以用求两个已排好序的数组中第k大数字来解决
求在两个已排好序的数组中的第k个数可以参考http://www.cnblogs.com/buptLizer/archive/2012/03/31/2427579.html
c++代码
1 class Solution { 2 public: 3 double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { 4 int m = nums1.size(); 5 int n = nums2.size(); 6 int total = m + n; 7 if(total & 1) 8 return findKth(nums1, 0, nums2, 0, total / 2 + 1) * 1.0; 9 else 10 return (findKth(nums1, 0, nums2, 0, total / 2) + findKth(nums1, 0, nums2, 0, total / 2 + 1)) / 2.0; 11 } 12 private: 13 int findKth(vector<int> &nums1, int i, vector<int> &nums2, int j, int k) 14 { 15 if(nums1.size() - i > nums2.size() - j) 16 return findKth(nums2, j, nums1, i, k); 17 if(nums1.size() == i) 18 return nums2[j + k - 1]; 19 if(k == 1) 20 return min(nums1[i], nums2[j]); 21 int pa = min(i + k / 2, (int)nums1.size()); 22 int pb = j + k - pa + i; 23 if(nums1[pa - 1] < nums2[pb - 1]) 24 return findKth(nums1, pa, nums2, j, k - pa + i); 25 else if(nums1[pa - 1] > nums2[pb - 1]) 26 return findKth(nums1, i, nums2, pb, k - pb + j); 27 else return nums2[pb - 1]; 28 } 29 };