[leetcode] Add Two Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46473791

本文深入探讨了LeetCode中关于两数相加问题的实现，通过实例展示了如何利用链表结构高效求解，并详细解释了代码逻辑与核心算法原理。

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From : https://leetcode.com/problems/add-two-numbers/

You are given two linked lists representing two non-negative numbers. 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.

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if(!l1) return l2;
        if(!l2) return l1;
        int carry=0, sum;
        ListNode* thead = new ListNode(0), *pre=thead;
        while(l1 && l2) {
            sum = l1->val + l2->val + carry;
            carry = sum>9;
            ListNode *node = new ListNode(sum%10);
            pre->next = node;
            pre = node;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1) {
            sum = l1->val + carry;
            carry = sum>9;
            ListNode *node = new ListNode(sum%10);
            pre->next = node;
            pre = node;
            l1 = l1->next;
        }
        while(l2) {
            sum = l2->val + carry;
            carry = sum>9;
            ListNode *node = new ListNode(sum%10);
            pre->next = node;
            pre = node;
            l2 = l2->next;
        }
        if(carry) {
            ListNode *node = new ListNode(1);
            pre->next = node;
        }
        pre = thead->next;
        delete thead;
        return pre;
    }
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if(!l1) return l2;
        if(!l2) return l1;
        int carry=0, sum, a, b;
        ListNode* thead = new ListNode(0), *pre=thead;
        while(l1 || l2) {
            a=b=0;
            if(l1) {a=l1->val; l1=l1->next; }
            if(l2) {b=l2->val; l2=l2->next; }
            sum = a + b + carry;
            carry = sum>9;
            ListNode *node = new ListNode(sum%10);
            pre->next = node;
            pre = node;
        }
        if(carry) pre->next = new ListNode(1);
        pre = thead->next;
        delete thead;
        return pre;
    }
};