【LeetCode】4. Median of Two Sorted Arrays

最新推荐文章于 2021-03-09 19:39:30 发布

原创最新推荐文章于 2021-03-09 19:39:30 发布 · 296 阅读

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本文介绍了一种求解两个已排序数组中位数的方法，通过合并两个数组并计算中位数，整体运行时间复杂度为O(log(m+n))。示例展示了不同情况下的中位数计算。

Description：

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

题目分析：按照归并排序的思路，先归并，再计算中间值。

Solution：

class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size();
int n = nums2.size();
vector<int> array = merge(nums1, nums2);
return ((double)(array[(m+n-1)/2]+array[(m+n)/2]))/2;
}
vector<int> merge(vector<int> A, vector<int> B)
{
vector<int> ret;
int m = A.size();
int n = B.size();
int i = 0;
int j = 0;
while(i < m && j < n)
{
if(A[i] <= B[j])
{
ret.push_back(A[i]);
i ++;
}
else
{
ret.push_back(B[j]);
j ++;
}
}
if(i == m)
{
while(j < n)
{
ret.push_back(B[j]);
j ++;
}
}
if(j == n)
{
while(i < m)
{
ret.push_back(A[i]);
i ++;
}
}
return ret;
}
};