LeetCode第四题（Median of Two Sorted Arrays）

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

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python代码

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        nums = []
        md = 0
        if len(nums1) == 0:
            nums = nums2
        elif len(nums2) == 0:
            nums = nums1
        elif nums1[len(nums1)-1] <= nums2[0]:        #第一个数组最后一位比第二个数组第一位小
            nums = nums1 + nums2
        elif nums1[0] >= nums2[len(nums2)-1]:        #第一个数组第一位比第二个数组最后一位大
            nums = nums2 + nums1
        else:
            p1 = 0
            p2 = 0
            for i in range(len(nums1)+len(nums2)):
                if p1 == len(nums1):
                    nums.append(nums2[p2])
                    p2 += 1
                elif p2 == len(nums2):
                    nums.append(nums1[p1])
                    p1 += 1
                else:
                    if nums1[p1] <= nums2[p2]:
                        nums.append(nums1[p1])
                        p1 += 1
                    else:
                        nums.append(nums2[p2])
                        p2 += 1
        if len(nums) % 2 == 0:    #偶数个
            md = (nums[int(len(nums)/2)-1] +     nums[int(len(nums)/2)]) / 2
        else:                     #奇数个
            md = nums[int(len(nums)/2)]
        return float(md)

PS：其实这个算法是最容易想到的算法，利用两个有序列表的特性，把两个列表整合成一个列表再找中位数，所以效率并不出众，下图足以说明：