2. Add Two Numbers

最新推荐文章于 2021-12-02 14:05:46 发布

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本文介绍如何解决LeetCode上的两数相加问题，使用链表存储两个非负整数并返回它们的和。提供了C++和Python两种语言的实现方案。

题目

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

分析：

从起点一直加进去，难度不大。

c++实现

/**
 * 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 == NULL) return l2;
        if(l2 == NULL) return l1;
        ListNode *p1, *p2, *pr = NULL, *phead = NULL;
        p1 = l1;
        p2 = l2;
       
       while(p1 != NULL || p2 != NULL){
           if(pr == NULL){
               //cout<< "new point"<<endl;
                pr = (ListNode*) malloc(sizeof(ListNode));
                phead = pr;
                pr->val = 0;
           } 
           else {
               //cout<< "next point"<<endl;
               pr->next = (ListNode*) malloc(sizeof(ListNode));
               if(pr->val >= 10){
                   pr->val -= 10;
                   pr->next->val = 1;
               }
               else
                    pr->next->val = 0;
                pr = pr->next;
           }
            
            if(p1 != NULL){
                pr->val += p1->val;
                p1 = p1->next;
            }
            if(p2 != NULL){
                pr->val += p2->val;
                p2 = p2->next;
            }
       }
       if(pr->val >= 10){//
           pr->next = (ListNode*) malloc(sizeof(ListNode));
           pr->val -= 10;
           pr->next->val = 1;
           pr = pr->next;
       }
       pr->next = NULL;
       return phead; 
    }
};

python 实现

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if l1 is None: return l2
        if l2 is None: return l1
        last_node = node = None
        muil = 0
        while l1 is not None and l2 is not None:
            cur_node, muil = self.addNodeValue(muil, l1.val+l2.val)
            if node is None:
                node =  cur_node
            if last_node != None:
                last_node.next = cur_node
            last_node = cur_node
            l1 = l1.next
            l2 = l2.next
            
        if l1 is None:
            l1 = l2
        while l1 is not None:
            if muil == 0:
                last_node.next = l1
                break
            cur_node, muil = self.addNodeValue(muil, l1.val)
            last_node.next = cur_node
            last_node = last_node.next
            l1 = l1.next
        if muil == 1:
            last_node.next = ListNode(muil)
        return node
    
    def addNodeValue(self, muil, val):
        num = muil + val
        if num < 10:
            return  ListNode(num), 0
        else:
            return ListNode(num-10), 1