2.两数相加

题目链接

题解

模拟加法的过程。

有个链表的技巧，如果再头指针可能发生变化的情况下，最好添加一个头结点，这样对于链表的许多操作不需要特判头指针了。

CPP

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        auto dummy = new ListNode(-1), cur = dummy;
        int t = 0;
        while(l1 || l2 || t){
            if(l1) t += l1->val, l1 = l1->next;
            if(l2) t += l2->val, l2 = l2->next;
            cur->next = new ListNode(t % 10);
            cur = cur->next;
            t /= 10;
        }

        return dummy->next;
    }
};

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        int t = 0;
        while(l1 != null || l2 != null || t != 0){
            if(l1 != null) {
                t += l1.val;
                l1 = l1.next;
            }
            if(l2 != null){
                t += l2.val;
                l2 = l2.next;
            }
            cur.next = new ListNode(t % 10);
            cur = cur.next;
            t /= 10;
        }
        
        return dummy.next;
    }
}