5 leetcode - Median of Two Sorted Arrays

Median of Two Sorted Arrays,两个有序数组的中位数

题目

英文:
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)).

中文:两个排好序的数组,找到两个数组的中间数,要求时间复杂度O(log (m+n))

举例:
nums1 = [1, 3] nums2 = [2] The median is 2.0
nums1 = [1, 2] nums2 = [3, 4] The median is (2 + 3)/2 = 2.5

直接合并两个数组，直至合并至中位数

class Solution(object):    def findMedianSortedArrays(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: float        """        len1,len2 = len(nums1),len(nums2)        if len1 == 0 and len2 == 0:            return 0        median = int((len1 + len2)/2)  #排序到中间位即可        target = []        index1 = index2 = 0        #如果偶数        while (index1 + index2) <= median and index1 < len1 and index2 <len2:            if nums1[index1] < nums2[index2]:                target.append(nums1[index1])                index1 += 1            else:                target.append(nums2[index2])                index2 += 1        #未取到中间位的数,便有一个list空了,list衔接        if (index1 + index2) <= median:            if index1 < len1:                target.extend(nums1[index1:])            if index2 < len2:                target.extend(nums2[index2:])         return target[median] if (len1 + len2)%2 else float(target[median] + target[median - 1])/2

二分查找两个数组的第k个数

详细解释见http://blog.csdn.net/lis_12/article/details/53128594

class Solution(object):    def findMedianSortedArrays(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: float        """        len1,len2 = len(nums1),len(nums2)        if (len1 + len2) & 1:  #奇数个            return self.fun(nums1,nums2,(len1 + len2)/2 + 1)        else:#偶数个            return float(self.fun(nums1,nums2,(len1 + len2)/2 + 1) + self.fun(nums1,nums2,(len1 + len2)/2))/2    def fun(self,nums1,nums2,k):        #如果一个list为空,肯定返回另外一个的第k个元素        if len(nums2) == 0:            return nums1[k - 1]        #保证nums1长度大于nums2        if len(nums2) > len(nums1):            return self.fun(nums2,nums1,k)        #取第一个元素,肯定是要比两个list的首个元素        if k == 1:            return min(nums1[0],nums2[0])        q = min(k / 2, len(nums2))#这里很容易,出错,防止越界.....        p = k - q        if nums1[p - 1] < nums2[q - 1]:            return self.fun(nums1[p:],nums2,k - p)        elif nums1[p - 1] > nums2[q - 1]:            return self.fun(nums1,nums2[q:],k - q)        else:            return nums1[p - 1]