【leetcode】4. Median of Two Sorted Arrays（Python & C++）

4. Median of Two Sorted Arrays

题目链接

4.1 题目描述：

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

4.2 解题思路：

思路一：首先处理特殊情况，如果其中一个数组为空，则根据长度可直接返回另一个数组的中位数。否则同时遍历两个数组，将其按照顺序合并到一个数组中，然后根据长度即可返回中位数。

4.3 C++代码：

1、思路一代码（82ms）：

class Solution93_1 {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
if (nums1.size() == 0 && nums2.size() == 0)
return 0;
if (nums1.size() == 0)
{
int l = nums2.size() / 2;
if (nums2.size() % 2 == 0)
return double(nums2[l - 1] + nums2[l]) / 2;
else
return double(nums2[l]);
}
if (nums2.size() == 0)
{
int l = nums1.size() / 2;
if (nums1.size() % 2 == 0)
return double(nums1[l - 1] + nums1[l]) / 2;
else
return double(nums1[l]);
}
int ll = (nums1.size() + nums2.size()) / 2;
int lu = (nums1.size() + nums2.size()) % 2;
int i = 0, j = 0;
vector<int>a;
while (i < nums1.size() && j < nums2.size())
{
if (nums1[i] <= nums2[j])
{
a.push_back(nums1[i]);
i++;
}
else
{
a.push_back(nums2[j]);
j++;
}
}
while (i < nums1.size())
{
a.push_back(nums1[i]);
i++;
}
while (j < nums2.size())
{
a.push_back(nums2[j]);
j++;
}
if (lu == 0)
return double(a[ll-1] + a[ll]) / 2;
else
return double(a[ll]);
}
};

4.4 Python代码：

1、思路一代码（118ms）：

class Solution(object):
def findMedianSortedArrays(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: float
"""
if len(nums1)==0 and len(nums2)==0:
return 0
if len(nums1)==0:
l = len(nums2) / 2
if len(nums2)% 2 == 0:
return float(nums2[l - 1] + nums2[l]) / 2
else:
return float(nums2[l])
if len(nums2)==0:
l = len(nums1) / 2
if len(nums1)% 2 == 0:
return float(nums1[l - 1] + nums1[l]) / 2
else:
return float(nums1[l])
ll = (len(nums1) + len(nums2)) / 2
lu = (len(nums1) + len(nums2)) % 2
i = 0
j = 0
a=[]
while i<len(nums1) and j<len(nums2):
if nums1[i] <= nums2[j]:
a.append(nums1[i])
i+=1
else:
a.append(nums2[j])
j+=1
while i<len(nums1) :
a.append(nums1[i])
i+=1
while j<len(nums2):
a.append(nums2[j])
j+=1
if lu == 0:
return float(a[ll-1]+a[ll])/2
else:
return float(a[ll])