Sorting a linked list in c++

别说谁变了你拦得住时间么 提交于 2019-12-03 21:39:57
sircodesalot

An easier way to slide through a linked list is like this:

for (node *current = head; current != nullptr; current = current->next) {
    // This will run through all of the nodes until we reach the end.
}

And to slide to the second to last item (ensuring that node->next exists) looks like this:

for (node *current = head; current->next != nullptr; current = current->next) {
    // Go through all of the nodes that have a 'next' node.
}

If you want to count how many items are in a linked list, you do something like this:

int count = 0;
for (node *current = head; current != nullptr; current = current->next) {
    count = count + 1;
}

So a selection type sort like you have above would look like this:

for (node *index = head; index->next != nullptr; index = index->next) {
  for (node *selection = index->next; selection != nullptr; selection = selection->next) {
    if (index->data > selection->data) {
      swap(index->data, selection->data);
    }
  }
}

Although sorting linked lists is generally not the best way to go (unless you're performing a merge).

the problem is you are looping till count and you are incrementing count in every run of the loop//remove line count++ to avoid remove infinite loop

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!