LeetCode 热题 100 —— 两数相加

原题链接

"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。

模拟运算

复杂度分析
时间复杂度 O(max(m,n))
空间复杂度 O(1)

解题思路

我们同时遍历两个链表，存在链表中同一位置的值直接相加，开辟一个carry用来存取进位制，若链表长度不一致则补0

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

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

            carry = sum / 10;
            if (l1)
                l1 = l1 -> next;
            
            if (l2)
                l2 = l2 -> next;
        }

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

        return head;
    }
};

/**
 * 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, 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);
        }
        return head;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) (head *ListNode) {
    var tail *ListNode
    carry := 0

    for l1 != nil || l2 != nil {
        n1, n2 := 0, 0
        if l1 != nil {
            n1 = l1.Val
            l1 = l1.Next
        }

        if l2 != nil {
            n2 = l2.Val
            l2 = l2.Next
        }

        sum := n1 + n2 + carry
        sum, carry = sum % 10, sum / 10

        if head == nil {
            head = &ListNode{Val: sum}
            tail = head
        } else {
            tail.Next = &ListNode{Val: sum}
            tail = tail.Next
        }
    }
    if carry > 0 {
        tail.Next = &ListNode{Val: carry}
    }
    return
}

# 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]:
        cur = dummy = ListNode()
        carry = 0
        while l1 or l2 or carry:
            carry += (l1.val if l1 else 0) + (l2.val if l2 else 0)
            cur.next = ListNode(carry % 10)
            carry //= 10
            cur = cur.next
            if l1: l1 = l1.next
            if l2: l2 = l2.next
        return dummy.next