Question

使用合并排序算法将每个长度为n的n个字符串列表排序为字典顺序。最坏的情况是这个计算的运行时间是？

我把这个问题作为家庭作业。我知道在O（nlogn）时间内合并排序排序。对于长度的词典顺序是n次nlogn？还是n ^ 2？

Answer 1

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 - 你可能会检测到哪一个只在最后一个字符上“更大”，你有O(nlogn)个比较在mergesort。

因此，您得到O(nlogn * n) = O(n^2 * logn)

Answer 2

但根据递归关系

T（n）= 2T（n / 2）+ O（m * n）

会的当m = n时，T（n）= 2T（n / 2）+ O（n ^ 2）

然后结果将是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

时间复杂度递归关系是

T（A，B）= 2T（A / 2，B）+ O（B ^ 2）

很明显树的高度是登录的。因此，时间复杂度为O（n ^ 2 * logn）。

合并排序词典排序的最坏情况运行时间？

4 个答案: