添加兩個數字作為鏈表

Question

我正在嘗試添加兩個數字（以相反順序的數字表示為鏈表）。 我在 C++ 中有一個解決方案

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        bool carry = 0;
        ListNode *head1 = l1; 
        ListNode *head2 = l2; 

        ListNode *head = nullptr;
        ListNode *curr = nullptr;

        while (head1 || head2 || carry) {
            // get value
            int val = (head1 ? head1->val : 0) +
                (head2 ? head2->val : 0) + carry;

            curr = new ListNode(val % 10);
            if (!head) head = curr;

            curr = curr->next;
            head1 = head1 ? head1->next : nullptr; 
            head2 = head2 ? head2->next : nullptr;
            carry = val / 10;
        }

        return head;
    }
};

出於某種原因，這只會返回長度為 1 的鏈表。 這不應該在第一次將 head 初始化為 curr，然后 curr 將繼續正確構建列表嗎？

Answer 1

正如 Ian4264 和 JaMiT 提到的，第一次進入循環時，你創建一個新節點並使curr指向它。 在此之后，您立即將curr設置為curr->next並且它指向某個任意位置。 您永遠不會將curr->next點設置為您將進行進一步分配的位置。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        bool carry = 0;
        ListNode *head1 = l1; 
        ListNode *head2 = l2; 

        ListNode *head = nullptr;
        ListNode *curr = nullptr;

        while (head1 || head2 || carry) {
            // get value
            int val = (head1 ? head1->val : 0) +
                (head2 ? head2->val : 0) + carry;


            if (!curr){
                //first time around the loop (curr is nullptr) 
                //set curr to point to a new node
                curr = new ListNode(val % 10);
            }
            else {
                //if current was pointing to some node already, then 
                //its's next is set to point to the new allocation and curr
                //is set to the last node (just allocated) - and thus 
                //current
                curr->next = new ListNode(val % 10);
                curr = curr->next;
            }
            if (!head) head = curr;

            head1 = head1 ? head1->next : nullptr; 
            head2 = head2 ? head2->next : nullptr;
            carry = val / 10;
        }

        return head;
    }
};