# 2020q3 Homework1 (quiz1)
## AA1和AA2

void add_entry(node_t **head, int new_value)
{
    node_t **indirect = head;

    node_t *new_node = malloc(sizeof(node_t));
    new_node->value = new_value;
    new_node->next = NULL;

    AA1;
    while (*indirect)
        indirect = &(*indirect)->next;
    AA2;
}

- ### AA1
如果這邊放```*indirect = new_node;```會把```head```設為```new_node```所以不可能。所以選擇檢查```new_node```的```assert(new_node);```
- ### AA2
這邊要把```new_node```放入linked list中的最後一個節點，所以選擇```*indirect = new_node;```
## BB1和BB2

node_t *swap_pair(node_t *head)
{
    for (node_t **node = &head; *node && (*node)->next; BB1) {
        node_t *tmp = *node;
        BB2;
        tmp->next = (*node)->next;
        (*node)->next = tmp;
    }
    return head;
}

- ### BB1
因為這邊要兩兩互換，而且下面```tmp```的變數會暫時除存一個節點的指標，所以在```for```的迴圈中要讓```node```的指標一次跳兩個
- ### BB2
這邊因為上面讓```tmp```存了```node```而下面又使```tmp->next = (*node)->next```所以BB2這行必須要使```tmp```或```node```改變，所以這邊```tmp = tmp->next;```或```(*node) = (*node)->next;```都可以
## CCC

node_t *reverse(node_t *head)
{
    node_t *cursor = NULL;
    while (head) {
        node_t *next = head->next;
        CCC;
        head = next;
    }
    return cursor;
}

- ### CCC
這邊因為要把linked list 做 reverse ，所以用```cursor```去當新的head並且邊移邊轉。```next```是用來保存```head->next```的節點，讓```head```轉過去後還能夠換到原本的```head->next```。

## swap_pair改寫
因為```head```只用用來設定```node```而已，所以如果讓```head```來代替```node```的工作則可改寫為：

void swap_pair(node_t **head){
    for(;*head && (*head)->next;head = &*=(*head)->next->next){
        node_t *tmp = *head;
        *head=(*head)->next;
        tmp->next = (*head)->next;
        (*head)->next = tmp;
    }

## reverse改寫
這邊就單純把```head```變成雙重指標，然後在最後的時候把```*head = cursor```

void reverse(node_t **head){
    node_t *cursor = NULL;
    while(*head){
        node_t *next = (*head)->next;
        (*head)->next = cursor;
        cursor = *head;
        *head = next;
    }
    *head = cursor;
}