# info2021-homework1
###### tags: `資訊科技產業面試`
contributed by <`bochengC>

### 總體初步檢討
#### 對interviewee
1. 不知道該如何問出延伸的問題，如何跟實務有連接
2. 用英文的時候，會明顯有斷點，無法連續輸出
3. 對語法不熟，`struct` 的分號，在結束後才想到
4. 無法用語言具體的有邏輯的將想法敘述出來
5. 不該照抄Leetcode的程式碼，應該直接如以下寫法

sturct ListNode{};

NodeList* function_name(ListNode* ){

}......

不用在那邊public之類的

#### 對interviewer
1. 詞彙不足，不知道怎麼跟interviewee互動
2. 沒辦法跟interviewee 討論想法

### easy1 1290. Convert Binary Number in a Linked List to Integer

:red_square:
:white_large_square: :

struct ListNode{
    int val;
    ListNode* next;
};

class Solution {
public:
    int getDecimalValue(ListNode* head) {
        int sum = 0 ;
        ListNode* pointer= head;
        while(pointer != NULL){
            sum = sum *2 ;
            sum = sum + pointer->val;
            pointer= pointer->next;
        }
        return sum;
    }
};

### easy2 1474. Delete N Nodes After M Nodes of a Linked List

class Solution {
public:
    ListNode* deleteNodes(ListNode* head, int m, int n) {
        ListNode* pointer=head;
        int remove = n;
        int keep = m-1 ; 
        while(pointer != NULL){
            if(keep != 0){
                pointer = pointer->next;
                keep--;
            }else if(remove != 0){
                if(pointer->next == NULL) break; 
                pointer->next = pointer->next->next;
                remove--;
            }else{
                keep = m;
                remove = n ;
            }
        }
        return head;
        
    }
};

### medium 1669. Merge In Between Linked Lists

class Solution {
public:
    ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {
        ListNode *head, *tmp; 
        head = list1 ;
        a--; 
        while(a>0){
            list1 = list1->next;
            a--;
            b--;
        } 
        tmp = list1;
        while(b> 0){
            tmp = tmp->next;
            b--;
        }
        list1 ->next = list2 ;
        while(list2->next){
            list2 = list2->next;
        }
        list2->next = tmp->next;
        return head;
    }
};