[英]Sorting a linked list in c++
I'm getting mad with infinite loop, what do you think is suitable solution? 我对无限循环感到生气,您认为什么是合适的解决方案?
void sorting () {
node * temphead = head;
node * tempnode = NULL;
for (int i=0; i<count; i++) {
for (int j=0; j<count-i; j++) {
if (temphead->data > temphead->next->data) {
tempnode = temphead;
temphead = temphead->next;
temphead->next = tempnode;
}
temphead=temphead->next;
count++;
}
}
}
I tried to increment count and use many conditions with while- before and after the for loop with no result 我试图增加计数并在for循环之前和之后使用while的许多条件,但没有结果
An easier way to slide through a linked list is like this: 在链接列表中滑动的更简单方法是这样的:
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: 滑到倒数第二个项目(确保
node->next
存在)如下所示:
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: 如果要计算链接列表中有多少项,请执行以下操作:
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: 所以像上面的选择类型排序看起来像这样:
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). 尽管对链接列表进行排序通常不是最好的方法(除非您正在执行合并)。
问题是您要循环直到计数,并且在每次循环运行中都递增计数//删除行计数++,以避免删除无限循环
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