LeetCode OJ : 2 Add Two Numbers

最新推荐文章于 2020-12-23 14:11:37 发布

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本文介绍了一种解决两数相加问题的方法，通过链表形式存储非负整数，数字以逆序方式存储。提供了C++实现的示例代码，并讨论了处理进位的细节。

Add Two Numbers

Total Accepted: 103820 Total Submissions: 487934 Difficulty: Medium

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) {
        ListNode *ListSum = NULL, *pNode = NULL, *tail = NULL;
        ListNode *pl1 = l1, *pl2 = l2;
        int sum, cnt = 0;
        
        while(pl1 != NULL || pl2 != NULL){
            if(pl1 == NULL) {                   //此时是因为链表pl2比较长
                sum = cnt + pl2->val;
            }
            else if(pl2 == NULL){               //此时是因为链表pl1比较长
                sum = cnt + pl1->val;
            }
            else if(pl1 != NULL && pl2 != NULL){
                sum = cnt + pl1->val + pl2->val;
            }
            
            cnt = 0;
            if(sum >= 10){
                cnt = 1;
                sum -= 10;
            }
            
            pNode = new struct ListNode(sum);
            if(ListSum == NULL){
                ListSum = pNode;
                tail = pNode;
            }
            else{
                tail->next = pNode;
                tail = pNode;
            }
            
            if(pl1 != NULL)pl1 = pl1->next;
            if(pl2 != NULL)pl2 = pl2->next;
        }
        
        if(cnt == 1){                           //考虑到最高位的两位进位
             sum = cnt;
             pNode = new struct ListNode(sum);
             if(ListSum == NULL){
                ListSum = pNode;
                tail = pNode;
             }
             else{
                tail->next = pNode;
                tail = pNode;
             }
        }
        
        return ListSum;
    }
};