mergesort

来源： https://stackoverflow.com/questions/9401019/writing-merge-sort-in-php

来源： https://stackoverflow.com/questions/63946865/implementing-bottom-up-merge-sort-utilizing-javas-parallelstreams

问题 I have been working on a templated implementation of a linked-list, on purpose, to reinvent the wheel, to stumble into just this type problem to help learn the subtle nuances of pointer to class instance handling. The problem I have stumbled into has to do with merging sublists where on the second merge (the first merge where sublists can have multiple nodes) fails where a prior class instance (either from split or mergesorted ) appears to go out of scope (which should not have any affect on

问题 So what I want to do is to sort an linked list containing only strings. To do so, I have 2 options. Option 1 - dynamically allocate an array with the same size as the linked list and the strings containing it also with the same size, copy the contents of the linked list into the array and sort it using qsort . Option 2 - implement a merge sort algorithm in order to sort it. One of the problems is will it cost more memory and time if I do option 2 over option 1 or the option is the better? My

问题 I'm trying to modify a common mergesort, for sorting a struct of strings. But I can't figure out what's wrong. A warning appears on every line with strcpy() & strcmp() that tells: warning: passing argument 1 of 'strcpy' makes pointer from integer without a cast [-Wint-conversion]| C:\Program Files (x86)\CodeBlocks\MinGW\include\string.h|45|note: expected 'char *' but argument is of type 'char'| And it recommends me to put * on char L[n1] and R[n2] But even if i declare char *L[n1] char *R[n2]

问题 I am currently trying to formulate a mergeSort mechanism for a singly linked list. Through research and finding consistent ideas about A) a merge sort being the best way to sort a singly linked list, and B) that these are the key components for performing such an operation, I have arrived at this following code. It almost works exactly as intended, but will only return all of the integers larger than the last inputted number. For example, inputting 7, 6, 5, 4, 3, 2, 1 will return 1, 2, 3, 4,

问题 I have gone through Merge sort algorithm. I understood the logic, but I didn't get why we have to copy the b[] array again into the a[] array. What we have entered in the b[] array are sorted numbers only, right? But, if we print the b[] array, we are getting unsorted array. Once we copy it into the a[] array, we are getting the correct output. Can anyone explain why we have to copy the b[] array to the a[] array? Algorithm:: void MergeSort(int low, int high) { if (low < high) { int mid =

问题 I have gone through Merge sort algorithm. I understood the logic, but I didn't get why we have to copy the b[] array again into the a[] array. What we have entered in the b[] array are sorted numbers only, right? But, if we print the b[] array, we are getting unsorted array. Once we copy it into the a[] array, we are getting the correct output. Can anyone explain why we have to copy the b[] array to the a[] array? Algorithm:: void MergeSort(int low, int high) { if (low < high) { int mid =

问题 I have gone through Merge sort algorithm. I understood the logic, but I didn't get why we have to copy the b[] array again into the a[] array. What we have entered in the b[] array are sorted numbers only, right? But, if we print the b[] array, we are getting unsorted array. Once we copy it into the a[] array, we are getting the correct output. Can anyone explain why we have to copy the b[] array to the a[] array? Algorithm:: void MergeSort(int low, int high) { if (low < high) { int mid =

问题 I came across the following implementation of the mergeSort algorithm: def merge_sort(x): merge_sort2(x,0,len(x)-1) def merge_sort2(x,first,last): if first < last: middle = (first + last) // 2 merge_sort2(x,first,middle) merge_sort2(x,middle+1,last) merge(x,first,middle,last) def merge(x,first,middle,last): L = x[first:middle+1] R = x[middle+1:last+1] L.append(999999999) R.append(999999999) i=j=0 for k in range(first,last+1): if L[i] <= R[j]: x[k] = L[i] i += 1 else: x[k] = R[j] j += 1 x =