Level: Hard

Remark : 最簡單 的 Hard Level 題吧

思路 : 原本想說可能會要格式化 小數點後幾位，結果也不用。

Source : LeetCode

Description :

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
Follow up: The overall run time complexity should be O(log (m+n)).
Example 1:
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:
Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
Example 3:
Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Example 4:
Input: nums1 = [], nums2 = [1]
Output: 1.00000
Example 5:
Input: nums1 = [2], nums2 = []
Output: 2.00000
Constraints:
nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        mrg = []
        for i in nums1:
            mrg.append(i)
        for j in  nums2:
            mrg.append(j)
        mrg.sort()
        if len(mrg)%2!=0:
            return mrg[int(len(mrg)/2)]
        else:
            return (mrg[int(len(mrg)/2)-1]+mrg[int(len(mrg)/2)]) /2 
        print(mrg)
        return 0.111
            
        

Status : Success
Screeshoot :

TODO :
Steve Merritt - 寫程式不再崩潰！介紹 5 個 Google 工程師都在用的好習慣