字典排序的合并排序最坏情况运行时间？

Question

使用合并排序算法将每个长度为 n 的 n 个字符串列表按字典顺序排序。该计算的最坏情况运行时间是？

我把这个问题作为作业。我知道合并排序在 O(nlogn) 时间内排序。对于长度的字典顺序，它是 n 乘以 nlogn 吗？或 n^2 ？

Answer 1

该算法的每次比较都是 O(n) [比较两个字符串是 O(n) 最坏的情况 - 您可能仅在最后一个字符上检测到哪个“更大”] ，您在合并排序中进行了 O(nlogn) 比较。

因此你得到O(nlogn * n) = O(n^2 * logn)

Answer 2

但根据递推关系

T(n) = 2T(n/2) + O(m*n)

将
当 m = n 时，T(n) = 2T(n/2) + O(n^2)

那么结果将是 O(n^2) 而不是 O(n^2logn)。

如果我错了请纠正我。

Answer 3

**answer is O(n^2logn)
  , 
we know Merge sort has recurrence form
T(n) = a T(n/b) + O(n)
in case of merge sort 
it is 
T(n) = 2T(n/2) + O(n) when there are n elements
but here the size of the total is not "n" but "n string of length n"
so a/c to this in every recursion we are breaking the n*n elements in to half
for each recursion as specified by the merge sort algorithm
MERGE-SORT(A,P,R)  ///here A is the array P=1st index=1, R=last index in our case it 
                      is n^2 
if P<R
then Q = lower_ceiling_fun[(P+R)/2]
      MERGE-SORT(A,P,Q)
      MERGE-SORT(A,Q+1,R)
      MERGE (A,P,Q,R)
MERGE(A,P,Q,R) PROCEDURE ON AN N ELEMENT SUBARRAY TAKES TIME O(N)
BUT IN OUR CASE IT IS N*N
SO A/C to this merge sort recurrence equation for this problem becomes
T(N^2)= 2T[(N^2)/2] + O(N^2)
WE CAN PUT K=N^2 ie.. substitute to solve the recurrence
T(K)= 2T(K/2) + O(K)
a/c to master method condition T(N)=A T(N/B) + O(N^d)
                               IF A<=B^d  then T(N)= O(NlogN)
therefore T(K) = O(KlogK)
substituting K=N^2
we get T(N^2)= O(n*nlogn*n)
       ie.. O(2n*nlogn)
         .. O(n*nlogn)

因此解决了

Answer 4

时间复杂度递推关系为

T(a,b)=2T(a/2,b)+O(b^2)

显然，树的高度将是 logn 。
因此时间复杂度为O(n^2*logn)。