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2. Add Two Numbers

2. Add Two Numbers

作者: YellowLayne | 来源:发表于2017-06-14 10:05 被阅读0次

1.描述

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

2.分析

3.代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* l = NULL;
    struct ListNode* p = NULL;
    struct ListNode* p1 = l1;
    struct ListNode* p2 = l2;
    int plus = 0;
    while (p1 != NULL && p2 != NULL) {
        struct ListNode* q = (struct ListNode*)malloc(sizeof (struct ListNode));
        q->val = (p1->val + p2->val + plus) % 10;
        q->next = NULL;
        plus = (p1->val + p2->val + plus) / 10;
        if (l == NULL) {
            l = p = q;
        }else {
            p->next = q;
            p = q;
        }
        p1 = p1->next;
        p2 = p2->next;
    }
    
    while (p1 != NULL) {
        struct ListNode* q = (struct ListNode*)malloc(sizeof (struct ListNode));
        q->val = (p1->val + plus) %10;
        q->next = NULL;
        plus = (p1->val + plus) / 10;
        p->next = q;
        p = q;
        p1 = p1->next;
    }
    
    while (p2 != NULL) {
        struct ListNode* q = (struct ListNode*)malloc(sizeof (struct ListNode));
        q->val = (p2->val + plus) %10;
        q->next = NULL;
        plus = (p2->val + plus) / 10;
        p->next = q;
        p = q;
        p2 = p2->next;
    }
    
    while (plus != 0) {
        struct ListNode* q = (struct ListNode*)malloc(sizeof (struct ListNode));
        q->val = plus % 10;
        q->next = NULL;
        plus = plus/10;
        p->next = q;
        p = q;
    }
    
    return l;
}

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