俩数相加算法题解C/C++/JAVA/Python
  • 2019-12-22 14:18
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  • 算法题解

题目:

给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。

如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外,这两个数都不会以 0 开头。

示例:

输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)

输出:7 -> 0 -> 8

原因:342 + 465 = 807

来源:力扣(LeetCode)

链接:https://leetcode-cn.com/problems/add-two-numbers

C解法:

struct ListNode *addTwoNumbers(struct ListNode *l1, struct ListNode *l2)
{
    struct ListNode *head = (struct ListNode *)malloc(sizeof(struct ListNode));
    head->val = 0;
    head->next = NULL;
    struct ListNode *curr = head;
    int sum = 0, carry = 0;
    while (l1 != NULL || l2 != NULL)
    {
        int x = (l1 == NULL) ? 0 : l1->val;
        int y = (l2 == NULL) ? 0 : l2->val;
        sum = carry + x + y;
        carry = sum / 10;
        curr->next = (struct ListNode *)malloc(sizeof(struct ListNode));
        curr = curr->next;
        curr->val = sum % 10;
        curr->next = NULL;
        if (l1 != NULL)
            l1 = l1->next;
        if (l2 != NULL)
            l2 = l2->next;
    }
    if (carry == 1)
    {
        curr->next = (struct ListNode *)malloc(sizeof(struct ListNode));
        curr = curr->next;
        curr->val = carry;
        curr->next = NULL;
    }
    return head->next;
}

C++解法:

class Solution
{
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        ListNode *head = new ListNode(0);
        ListNode *curr = head;
        int sum = 0, carry = 0;
        while (l1 != NULL || l2 != NULL)
        {
            int x = (l1 == NULL) ? 0 : l1->val;
            int y = (l2 == NULL) ? 0 : l2->val;
            sum = carry + x + y;
            carry = sum / 10;
            curr->next = new ListNode(sum % 10);
            curr = curr->next;
            if (l1 != NULL)
                l1 = l1->next;
            if (l2 != NULL)
                l2 = l2->next;
        }
        if (carry == 1)
        {
            curr->next = new ListNode(carry);
            curr = curr->next;
        }
        return head->next;
    }
};

JAVA解法:

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode curr = head;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int x = (l1 == null) ? 0 : l1.val;
            int y = (l2 == null) ? 0 : l2.val;
            int sum = carry + x + y;
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);
            curr = curr.next;
            if (l1 != null)
                l1 = l1.next;
            if (l2 != null)
                l2 = l2.next;
        }
        if (carry > 0) {
            curr.next = new ListNode(carry);
        }
        return head.next;
    }
}

Python3解法:

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = ListNode(0)
        curr = head
        carry = 0
        while l1 or l2 :
            x = l1.val if l1 else 0
            y = l2.val if l2 else 0
            sum = carry + x + y
            carry = sum // 10
            curr.next = ListNode(sum % 10)
            curr = curr.next
            if l1 :
                l1 = l1.next
            if l2 :
                l2 = l2.next
        if carry == 1:
            curr.next = ListNode(carry)
        return head.next
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