文章目录

• • @[toc]
  • • 题目描述
    • 样例输入输出与解释
    • • 样例1
      • 样例2
    • 提示
    • Python实现
    • • 二分查找
      • 划分数组

个人主页：丷从心·

系列专栏：LeetCode

刷题指南：LeetCode刷题指南

---

题目描述

• 给定两个大小分别为m和n的正序（从小到大）数组nums1和nums2，请找出并返回这两个正序数组的中位数
• 算法的时间复杂度应该为O(log (m+n))

---

样例输入输出与解释

样例1

• 输入：nums1 = [1,3]，nums2 = [2]
• 输出：2.00000
• 解释：合并数组 = [1,2,3]，中位数2

样例2

• 输入：nums1 = [1,2]，nums2 = [3,4]
• 输出：2.50000
• 解释：合并数组 = [1,2,3,4]，中位数(2 + 3) / 2 = 2.5

---

提示

• nums1.length == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• -10^6 <= nums1[i], nums2[i] <= 10^6

---

Python实现

二分查找

class Solution:def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:def getKthElement(k):index1, index2 = 0, 0while True:# 特殊情况if index1 == m:return nums2[index2 + k - 1]if index2 == n:return nums1[index1 + k - 1]if k == 1:return min(nums1[index1], nums2[index2])# 正常情况new_index1 = min(index1 + k // 2 - 1, m - 1)new_index2 = min(index2 + k // 2 - 1, n - 1)pivot1, pivot2 = nums1[new_index1], nums2[new_index2]if pivot1 <= pivot2:k -= new_index1 - index1 + 1index1 = new_index1 + 1else:k -= new_index2 - index2 + 1index2 = new_index2 + 1m, n = len(nums1), len(nums2)total_length = m + nif total_length % 2 == 1:return getKthElement((total_length + 1) // 2)else:return (getKthElement(total_length // 2) + getKthElement(total_length // 2 + 1)) / 2	

划分数组

class Solution:def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:if len(nums1) > len(nums2):return self.findMedianSortedArrays(nums2, nums1)inf = 10 ** 6m, n = len(nums1), len(nums2)left, right = 0, mwhile left <= right:i = (left + right) // 2j = (m + n + 1) // 2 - inums_im1 = (-inf if i == 0 else nums1[i - 1])nums_i = (inf if i == m else nums1[i])nums_jm1 = (-inf if j == 0 else nums2[j - 1])nums_j = (inf if j == n else nums2[j])if nums_im1 <= nums_j:left = i + 1median1, median2 = max(nums_im1, nums_jm1), min(nums_i, nums_j)else:right = i - 1return (median1 + median2) / 2 if (m + n) % 2 == 0 else median1

---