C++刷題筆記：鏈表

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兩數相加

考慮創建啞節點dummy，使用dummy->next表示真正的頭節點，避免空邊界. 時間複雜度 $O(n)$

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* first = new ListNode(0);  // 啞結點
        ListNode* last = first;
        int carry = 0;
        while(l1 or l2 or carry){
            // 相加
            int bitsum = 0;
            if(l1){
                bitsum += l1->val;
                l1 = l1->next;
            }
            if(l2){
                bitsum += l2->val;
                l2 = l2->next;
            }
            if(carry){
                bitsum += carry;
            }
            // 結果
            int digit = bitsum % 10;
            carry = bitsum / 10;
            // 鏈表存儲
            ListNode* node = new ListNode(digit);
            last->next = node;
            last = node;
        }
        last = first->next;
        delete first;
        return last;
    }
};

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反轉鏈表輸出

struct ListNode {
      int val;
      struct ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
};

方法一：使用std::reverse()反轉鏈表的輸出vector

class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
	    std::vector<int> ret;
        while(head!=nullptr){
            ret.push_back(head->val);
            head= head->next;
        }
        //再開闢內存，逆序遍歷
        //for(auto iter=arr.rbegin(); iter<arr.rend();iter++){
        //    ret.push_back(*iter);
        //}
        std::reverse(ret.begin(), ret.end());
        return ret;
    }
};

方法二：直接反轉鏈表，改變指針指向

class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        ListNode* newHead  = nullptr;
        ListNode* node;  // 存放刪除的結點，也是新插入的結點
        while(head){
            node = head;            // 保存刪除
            head = head->next;      // 更新頭
            node->next = newHead;   // 重新插入
            newHead = node;         // 更新頭
        }
        
        std::vector<int> ret;
        while(newHead){
            ret.push_back(newHead->val);
            newHead = newHead->next;
        }
        return ret;
    }
};