2_两数相加

【问题描述】

给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照逆序的方式存储的，并且每个节点只能存储一位数字。请你将两个数相加，并以相同形式返回一个表示和的链表。你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例：
输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

解：

【算法设计思想】

注意理清题意，结合链表通常在尾部进行数据修改的特点，所以才会有题目中的“每位数字都是按照逆序的方式存储的”。在这里，要考虑到进位问题，大体的思路就是先分配一个空链表，设立头、尾两个指针，利用carry变量来记录进位的数字，模拟手算过程。在遍历过程结束以后，即计算到最高位时，要注意是否会进位。例如，566+468=1034，这里应该再分配一个内存空间来存储多余的那个“1”。

【算法描述】

C语言：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *head = NULL, *tail = NULL;
    int carry = 0;
    while (l1 || l2) {
        int n1 = l1 ? l1->val : 0;
        int n2 = l2 ? l2->val : 0;
        int sum = n1 + n2 + carry;
        if (!head) {
            head = tail = malloc(sizeof(struct ListNode));
            tail->val = sum % 10;
            tail->next = NULL;
        } else {
            tail->next = malloc(sizeof(struct ListNode));
            tail->next->val = sum % 10;
            tail = tail->next;
            tail->next = NULL;
        }
        carry = sum / 10;
        if (l1) {
            l1 = l1->next;
        }
        if (l2) {
            l2 = l2->next;
        }
    }
    if (carry > 0) {
        tail->next = malloc(sizeof(struct ListNode));
        tail->next->val = carry;
        tail->next->next = NULL;
    }
    return head;
}

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 head = null;
        ListNode tail = null;
        int carry = 0;
        while(l1 != null || l2 != null){
            int n1 = (l1 != null)? l1.val : 0;
            int n2 = (l2 != null)? l2.val : 0;
            int sum = n1 + n2 + carry;
            if(head == null){
                head = tail = new ListNode(sum % 10);
            }else{
                tail.next = new ListNode(sum % 10);
                tail = tail.next;
            }

            carry = sum / 10;

            if(l1 != null){
                l1 = l1.next;
            }

            if(l2 != null){
                l2 = l2.next;
            }


        }

        if(carry > 0){
            tail.next = new ListNode(carry);
            tail = tail.next;
        }

        return head;
    }
}

Python：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(
        self, l1: Optional[ListNode], l2: Optional[ListNode]
    ) -> Optional[ListNode]:
        head = None
        tail = None
        carry = 0
        while l1 != None or l2 != None:
            n1 = l1.val if (l1 != None) else 0
            n2 = l2.val if (l2 != None) else 0
            sum = n1 + n2 + carry

            if head == None:
                head = tail = ListNode(sum % 10)
            else:
                tail.next = ListNode(sum % 10)
                tail = tail.next

            carry = sum // 10

            if l1 != None:
                l1 = l1.next
            if l2 != None:
                l2 = l2.next

        if carry > 0:
            tail.next = ListNode(carry)
            tail = tail.next

        return head

C++:

/**
 * 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) {
        ListNode* head = nullptr;
        ListNode* tail = nullptr;
        int carry = 0;
        while (l1 != nullptr || l2 != nullptr) {
            int n1 = (l1 != nullptr) ? l1->val : 0;
            int n2 = (l2 != nullptr) ? l2->val : 0;
            int sum = n1 + n2 + carry;

            if (head == nullptr) {
                head = tail = new ListNode(sum % 10);
            } else {
                tail->next = new ListNode(sum % 10);
                tail = tail->next;
            }

            carry = sum / 10;

            if (l1 != nullptr) {
                l1 = l1->next;
            }

            if (l2 != nullptr) {
                l2 = l2->next;
            }
        }

        if (carry > 0) {
            tail->next = new ListNode(carry);
            tail = tail->next;
        }

        return head;
    }
};