leetcode Median of Two Sorted Arrays

本文介绍了一种使用二分查找法找到两个有序数组中位数的有效方法,通过调整切割点确保最终中位数的正确计算,并提供了一种更简单的寻找第k个元素的方法。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

There are two sorted arrays A and B 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)).

Best Solution by Binary Search:

Before writting codes,  we should understand the following things:

  1. The median of any ascending array is (num[ceiling((len-1)/2)]+num[ceiling(len/2)])/2. Mark cut (l,r)=(num[ceiling((len-1)/2)],num[ceiling(len/2)])/2). 
  2. For two sorted arrays, the corresponding cuts are cut1(l1,r1), cut2(l2,r2). If l1<=r2 and l2<=r1, the final median is (max(l1,l2)+min(r1,r2))/2
  3. However, item 2 won't always happen. If we adjust cut1 to one direction and cut2 to opposite direction until l1<=r2 and l2<=r1, we could stil find the final median (max(l1,l2)+min(r1,r2))/2
  4. Initially, cut1(l1,r1)=(num[ceiling((len1-1)/2)],num[ceiling(len1/2)])/2), cut2(l2,r2)=(num[ceiling((len2-1)/2)],num[ceiling(len2/2)])/2), final merged array cut is (num[ceiling((len1+len2-1)/2)],num[ceiling((len1+len2)/2)])/2). The question is that if updated (l1,r1) is (num[ceiling((len1-1-x)/2)],num[ceiling((len1-x)/2)])/2), what's updated cut2 to ensure (l1,r1), (l2,r2) could find the final median? The answer is updated cut2 = (num[ceiling((len2+x-1)/2)],num[ceiling((len2+x)/2)])/2), which could be understanded as move x elements from array1 to array2 but the final median won't change
  5. Consider some corner cases:
  6. So the task becomes how to find appropriate cut index (ceiling((len1-1-x)/2), ceiling((len1-x)/2)) in order to make l1<=r2 and l2<=r1, we could do binary search ref to https://blog.youkuaiyun.com/taoqick/article/details/23260157. Understand the range of start+off is closed interval [0, len]
  7. (ceiling((len-1)/2), ceiling((len1)/2)) and (ceiling((len1)/2), ceiling((len1+1)/2)) are different and look like walking half step. It seems that we move left leg with one step and draw close the other leg. In order to support the half step and ceiling((len1-1)/2 ranging in closed interval [-1, len1-1], the search range for len1 should be closed [0,2len1]. We could finish the codes now.
import sys
class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        n1, n2 = len(nums1), len(nums2)
        if (n1 < n2):
            return self.findMedianSortedArrays(nums2, nums1)
        #You may assume nums1 and nums2 cannot be both empty
        left, right = 0, n2*2
        l1_i,r1_i,l2_i,r2_i = 0,0,0,0
        while (left <= right):

            mid = left+(right-left)//2 #range from [0, 2*n2]

            l2_i,r2_i = (mid-1)//2, mid//2 #l2_i ranges from [-1, n2-1], r2_i ranges from [0, n2]
            l1_i,r1_i = (n1+n2-mid-1)//2, (n1+n2-mid)//2 #l1_i ranges from [-1, n1-1], r1_i ranges from [0, n1]

            l2, r2 = nums2[l2_i] if l2_i >= 0 else -sys.maxsize, nums2[r2_i] if r2_i < n2 else sys.maxsize
            l1, r1 = nums1[l1_i] if l1_i >= 0 else -sys.maxsize, nums1[r1_i] if r1_i < n1 else sys.maxsize

            if (l1 > r2):
                left = mid+1
            elif (l2 > r1):
                right = mid-1
            else:
                return (max(l1, l2) + min(r1, r2)) / 2

        return (max(l1,l2) + min(r1,r2)) / 2

s = Solution()
print(s.findMedianSortedArrays([1,3],[2,4,5]))
print(s.findMedianSortedArrays([3,4,5,6],[1,2,3]))
print(s.findMedianSortedArrays([1,2],[3]))
print(s.findMedianSortedArrays([1,2],[3,4]))
print(s.findMedianSortedArrays([2,3,4],[1]))

 

Another solution:

The method of finding the kth number is easier than http://leetcode.com/2011/03/median-of-two-sorted-arrays.html . 

Take notice of A[pa - 1] instead of A[pa] because pa >= 1

class Solution {
 public:
  double findKth(int A[], int m, int B[], int n, int k) {
    if (m > n)
      return findKth(B, n, A, m, k);
    if (m == 0)
      return B[k-1];
    if (k == 1)
      return min(A[0], B[0]);
    int pa = min(m, k/2);
    int pb = k - pa;
    if (A[pa - 1] < B[pb - 1])
      return findKth(A + pa, m - pa, B, n , k - pa);
    else if (A[pa - 1] > B[pb - 1])
      return findKth(A, m, B + pb, n - pb, k - pb);
    else
      return A[pa - 1];
  }
  double findMedianSortedArrays(int A[], int m, int B[], int n) {
    int total = m + n;
    if (total % 2 == 0)
      return (findKth(A, m, B, n, total/2) + findKth(A, m, B, n, total/2 + 1))/2;
    else
      return findKth(A, m, B, n, total/2 + 1);
  }
};

 

 

 

 

 

内容概要:文章基于4A架构(业务架构、应用架构、数据架构、技术架构),对SAP的成本中心和利润中心进行了详细对比分析。业务架构上,成本中心是成本控制的责任单元,负责成本归集与控制,而利润中心是利润创造的独立实体,负责收入、成本和利润的核算。应用架构方面,两者都依托于SAP的CO模块,但功能有所区分,如成本中心侧重于成本要素归集和预算管理,利润中心则关注内部交易核算和获利能力分析。数据架构中,成本中心与利润中心存在多对一的关系,交易数据通过成本归集、分摊和利润计算流程联动。技术架构依赖SAP S/4HANA的内存计算和ABAP技术,支持实时核算与跨系统集成。总结来看,成本中心和利润中心在4A架构下相互关联,共同为企业提供精细化管理和决策支持。 适合人群:从事企业财务管理、成本控制或利润核算的专业人员,以及对SAP系统有一定了解的企业信息化管理人员。 使用场景及目标:①帮助企业理解成本中心和利润中心在4A架构下的运作机制;②指导企业在实施SAP系统时合理配置成本中心和利润中心,优化业务流程;③提升企业对成本和利润的精细化管理水平,支持业务决策。 其他说明:文章不仅阐述了理论概念,还提供了具体的应用场景和技术实现方式,有助于读者全面理解并应用于实际工作中。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值