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

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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：

1. 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 iand jif nums1[i] <= nums2[j]:
        a.append(nums1[i])
        i+=1
        else:
        a.append(nums2[j])
        j+=1
    while i1        
    while j1        
    if lu == 0:
        return float(a[ll-1]+a[ll])/2
    else:
        return float(a[ll])

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