[英]Merging two linked lists
I have a task where I have to combine 2 linked list in a function in ascendent order without creating a new list.我有一个任务,我必须按升序组合 function 中的 2 个链表而不创建新列表。 Something like this: [1.3-->3.2-->4.6-->7.0]
& [2.7-->2.9-->5.1]
-> [1.3-->2.7-->2.9-->3.2-->4.6-->5.1-->7.0]
.像这样: [1.3-->3.2-->4.6-->7.0]
& [2.7-->2.9-->5.1]
-> [1.3-->2.7-->2.9-->3.2-->4.6-->5.1-->7.0]
。 I have the whole code but the monitoring program keep reporting the same: MEMORY LEAK
.我有完整的代码,但监控程序一直报告相同: MEMORY LEAK
。
This is my code (only the function part, and the struct
definition):这是我的代码(只有 function 部分和struct
定义):
typedef struct _listelem {
double data;
struct _listelem *next;
} listelem;
listelem *merge(listelem *a, listelem *b) {
listelem *lemarado = NULL;
listelem *mozgo = a;
listelem *elol2 = b;
lemarado = mozgo;
if (elol2 == NULL) // if the list is zero
return a;
else
if (mozgo == NULL)
return b;
if (mozgo->next == NULL) { //if the first list has only one element
while (elol2 != NULL) {
listelem *vege;
vege = (listelem *)malloc(sizeof(listelem));
vege->data = elol2->data;
lemarado->next = vege;
free(vege);
lemarado = lemarado->next;
elol2 = elol2->next;
}
lemarado->next = NULL;
return a;
}
mozgo = mozgo->next; // I am maintaining a pointer which points to the next data, and a pointer
while (mozgo != NULL) { // Which points to the data before, and I stick an element between them
if (elol2 == NULL) {
while (mozgo != NULL)
mozgo = mozgo->next;
while (lemarado != NULL)
lemarado = lemarado->next;
return a;
}
if (mozgo->data > elol2->data) {
listelem *hozzafuz;
hozzafuz = (listelem *)malloc(sizeof(listelem));
hozzafuz->data = elol2->data;
lemarado->next = hozzafuz;
hozzafuz->next = mozgo;
elol2 = elol2->next;
lemarado = lemarado->next;
} else {
mozgo = mozgo->next;
lemarado = lemarado->next;
}
if (mozgo == NULL) {
while (elol2 != NULL) {
listelem *vege;
vege = (listelem *)malloc(sizeof(listelem));
vege->data = elol2->data;
lemarado->next = vege;
free(vege);
lemarado = lemarado->next;
elol2 = elol2->next;
}
lemarado->next = NULL;
}
}
return a;
}
There is no need to allocate of free list nodes, you are just supposed to merge the sorted lists into a single list.无需分配空闲列表节点,您只需将已排序的列表合并为一个列表即可。 Using non-English variable names is confusing for people who do not speak Hungarian, which I'm afraid is very common.使用非英语变量名会让不会说匈牙利语的人感到困惑,这恐怕很常见。
Here is a modified version:这是修改后的版本:
typedef struct _listelem {
double data;
struct _listelem *next;
} listelem;
listelem *merge(listelem *a, listelem *b) {
listelem *head = NULL;
listelem *tail = NULL;
if (b == NULL)
return a;
else
if (a == NULL)
return b;
if (a->data <= b->data) {
head = a;
a = a->next;
} else {
head = b;
b = b->next;
}
tail = head;
while (a != NULL && b != NULL) {
if (a->data <= b->data) {
tail->next = a;
a = a->next;
} else {
tail->next = b;
b = b->next;
}
tail = tail->next;
}
if (a == NULL)
tail->next = b;
else
tail->next = a;
return head;
}
If you master double pointers, the above code can be simplified as:如果掌握了双指针,上面的代码可以简化为:
listelem *merge(listelem *a, listelem *b) {
listelem *head = NULL;
for (listelem **link = &head;; link = &(*link)->next) {
if (a == NULL) {
*link = b;
break;
}
if (b == NULL) {
*link = a;
break;
}
if (a->data <= b->data) {
*link = a;
a = a->next;
} else {
*link = b;
b = b->next;
}
}
return head;
}
struct thing *merge(struct thing *one,struct thing *two)
{
struct thing *result;
struct thing **pp; // will always point to the _pointer_ that will be assigned the next node.
result=NULL;
for(pp = &result; one && two; pp = &(*pp)->next) {
if(one->value <= two->value) {
*pp = one; one = one->next;
}
else {
*pp = two; two = two->next;
}
}
// When we get here, one and/or two will be NULL
*pp = (one) ? one : two;
return result;
}
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