Leetcode 2 两数相加

一、题目

  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 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。

  可以假设除了数字 0 之外,这两个数都不会以 0 开头。

  示例 1:

    输入:l1 = [2,4,3], l2 = [5,6,4]
    输出:[7,0,8]
    解释:342 + 465 = 807.

二、解法

  以逆序链表的方式存储数字,非常适合于两数相加。这时候,链表头代表数字的最低位,只需从头向尾遍历,依次相加进位即可。

       以最短的数的链表为终点,然后再将较长的数的后续位加入新的链表中。

  Python版:

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class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        upper_val = 0
        dummy = ListNode(0)
        cur = dummy
        while l1 and l2:
            value = l1.val + l2.val + upper_val
            upper_val = value // 10
            tmp_node = ListNode(value % 10)
            cur.next = tmp_node
            cur = tmp_node
            l1, l2 = l1.next, l2.next
 
        remain_l = l1 if l1 else l2
        while remain_l:
            value = remain_l.val + upper_val
            upper_val = value // 10
            tmp_node = ListNode(value % 10)
            cur.next = tmp_node
            cur = tmp_node
            remain_l = remain_l.next
 
        if upper_val > 0:
            tmp_node = ListNode(upper_val)
            cur.next = tmp_node
 
        return dummy.next

       C语言版:

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# include <stdlib.h>
 
struct ListNode {
    int val;
    struct ListNode *next;
};
 
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    struct ListNode* l = NULL;
    struct ListNode* cur = NULL;
    // struct ListNode* tmp = NULL;
    l = malloc(sizeof(struct ListNode));
    l->val = 0;
    l->next = NULL;
    cur = l;
 
    int value = 0;
    int upperVal = 0;
    while (l1 != NULL && l2 != NULL) {
        value = l1->val + l2->val + upperVal;
        upperVal = value / 10;
        value = value % 10;
         
        struct ListNode* tmp_node = malloc(sizeof(struct ListNode));
        tmp_node->val = value;
        tmp_node->next = NULL;
        cur->next = tmp_node;
        cur = cur->next;
 
        l1 = l1->next;
        l2 = l2->next;
    }
 
    if (l1 == NULL) {
        while (l2 != NULL ) {
            value = l2->val + upperVal;
            upperVal = value / 10;
            l2->val = value % 10;
            cur->next = l2;
            cur = cur->next;
            l2 = l2->next;
        }
    }
    if (l2 == NULL) {
        while (l1 != NULL ) {
            value = l1->val + upperVal;
            upperVal = value / 10;
            l1->val = value % 10;
            cur->next = l1;
            cur = cur->next;
            l1 = l1->next;
        }
    }
    if (upperVal > 0) {
        struct ListNode* node = malloc(sizeof(struct ListNode));
        node->val = upperVal;
        node->next = NULL;
        cur->next = node;
    }
    return l->next;
}
posted @   LeonYi  阅读(60)  评论(0编辑  收藏  举报
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