leetcode:2号题 两数相加

两数相加

解上述题，采用同样的思路。
go实现对比c++实现，结果如下图所示：

image.png

C++实现

/**
 * 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) {
        //注意：1、链表是逆序的，返回链表也是逆序
        //        按输入两链表顺序加，有进位的话，就需要在后一个节点加上。
        //      2、最后，需要保存链表头，同时还需要一个处理链表。
        ListNode *head = new ListNode(-1);
        ListNode *node = head;
        int carry = 0;
        while(l1 || l2) {
            int val1 = 0, val2 = 0;
            if (l1 != nullptr) {
                val1 = l1->val;
                l1 = l1->next;
            }
            if (l2 != nullptr) {
                val2 = l2->val;
                l2 = l2->next;
            }
            int sum = val1 + val2 + carry;
            carry = sum / 10;
            node->next = new ListNode(sum % 10);
            node = node->next;
        }
        if(carry != 0) {
            node->next = new ListNode(carry);
        }
        return head->next;
    }
};

GO实现

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    ans := new(ListNode)
    curr := ans
    carry := 0
    for (l1 != nil || l2 != nil) {
        val1,val2 := 0,0
        if l1 != nil {
            val1 = l1.Val
            l1 = l1.Next
        }
        if l2 != nil {
            val2 = l2.Val
            l2 = l2.Next
        }
        sum := val1 + val2 + carry
        carry = sum / 10
        temp := new(ListNode)
        temp.Val = sum % 10
        temp.Next = nil
        curr.Next = temp
        curr = curr.Next
    }
    if carry != 0 {
        temp := new(ListNode)
        temp.Val = carry
        temp.Next = nil
        curr.Next = temp
    }
    return ans.Next
}