Leetcode之Add Two Numbers解法

最新推荐文章于 2025-02-27 23:12:53 发布

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本文介绍了一种解决两数相加问题的链表实现方法。通过定义单链表结构，采用迭代方式处理两个链表数值的相加运算，并妥善处理进位问题。文章详细展示了如何在链表尾部添加新节点以处理最后的进位情况。

/**
 * 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) {
        int carry = 0;
        ListNode *l1_temp = l1, *l2_temp = l2, *head ,*head_temp;
        head_temp = head = (ListNode*) malloc(sizeof(ListNode));
        head -> next = NULL;
        while(l1_temp != NULL || l2_temp != NULL){
        // (l1_temp != NULL && l2_temp != NULL) 写法复杂
            head_temp -> next = (ListNode*) malloc(sizeof(ListNode));
            head_temp = head_temp -> next;
            head_temp -> next = NULL;
            int x = (l1_temp == NULL) ? 0 : l1_temp -> val;
            int y = (l2_temp == NULL) ? 0 : l2_temp -> val;
            head_temp -> val = (x + y + carry) % 10;
            carry = (x + y + carry) / 10;
            if(l1_temp != NULL) l1_temp = l1_temp -> next;
            if(l2_temp != NULL) l2_temp = l2_temp -> next;
        }
        // easy to forget
        if(carry){
            head_temp -> next = new ListNode(carry);
        }
        return head -> next;
    }
};