LeetCode 4. Median of Two Sorted Arrays

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 };

 

posted @ 2017-10-24 23:35  cp_cnblogs  阅读(118)  评论(0)  编辑  收藏  举报