LeetCode题解--2. Add Two Numbers

链接

LeetCode题目： https://leetcode.com/problems/add-two-numbers/

GitHub代码 ：https://github.com/gatieme/LeetCode/tree/master/002-AddTwoNumbers

CSDN题解：http://blog.csdn.net/gatieme/article/details/50809402

描述

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

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这道题其实是大数相加的处理，没什么难度，但需要注意以下几点：

• 因为存储是反过来的，即数字342存成2->4->3，所以要注意进位是向后的；

• 链表l1和l2长度可能不同，因此要注意处理某个链表剩余的高位；

• 最高位如果有进位，那么需要新增一个结点存储最高进位

• 每一位都是0~9的数字，不会超过一个位

思路

链表单独进位处理

C++

思路非常简单，利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;

#define DEBUG



#ifdef DEBUG
class ListNode
{
public :
    ListNode( )
    :val(0), next(NULL)
    {
    }

    ListNode(int x)
    :val(x), next(NULL)
    {
    }

    int val;
    ListNode *next;

};

#endif




/** * 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)
    {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.

        if((l1 == NULL) && (l2 == NULL))
        {
            return NULL;
        }
        int sum = 0;
        int carry = 0;     // 进位
        int current = 0;

        ListNode *p = new ListNode(0);
        ListNode *start = p;

        sum = l1->val + l2->val + carry;
        carry = sum / 10;

        p->val = sum % 10;

        l1 = l1->next;
        l2 = l2->next;

        while(l1 != NULL && l2 != NULL)
        {
            sum = l1->val + l2->val + carry;
            carry = sum / 10;

            p->next = new ListNode((int)sum % 10);
            p = p->next;

            l1 = l1->next;
            l2 = l2->next;
        }

        // 处理多出的位，因为两个链表可能不一样长度
        if(l1 != NULL)
        {
            while(l1 != NULL)
            {
                sum = l1->val + carry;
                carry = sum / 10;

                p->next = new ListNode((int)sum % 10);
                p = p->next;

                l1 = l1->next;
            }
        }
        else if(l2 != NULL)
        {
            while(l2 != NULL)
            {
                sum = l2->val + carry;
                carry = sum / 10;

                p->next = new ListNode((int)sum % 10);
                p = p->next;

                l2 = l2->next;
            }
        }

        if(carry != 0)
        {
            p->next = new ListNode((int)carry);
            p = p->next;
        }

        return start;
    }
};


#ifdef DEBUG

int main( int argc, char *argv[] ) { ListNode l1[5], l2[5]; ListNode *pl1 = &l1[0], *pl2 = &l2[0]; // Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) // Output: 7 -> 0 -> 8 l1[0].val = 2;l1[0].next = &l1[1]; l1[1].val = 4;l1[1].next = &l1[2]; l1[2].val = 7;l1[2].next = NULL; l2[0].val = 5;l2[0].next = &l2[1]; l2[1].val = 6;l2[1].next = &l2[2]; l2[2].val = 7;l2[2].next = NULL; Solution solu = Solution(); ListNode *currNode = solu.addTwoNumbers(pl1, pl2); std::cout <<currNode->val; while(currNode->next != NULL) { currNode = currNode->next; std::cout <<" -> " <<currNode->val; } return EXIT_SUCCESS; } #endif

Python

#!/usr/bin/env python
#coding=utf-8
#
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    # @return a ListNode
    def addTwoNumbers(self, l1, l2):
        ret = ListNode(0)
        cur = ret

        sum = 0
        while True:
            if l1 != None:
                sum += l1.val
                l1 = l1.next
            if l2 != None:
                sum += l2.val
                l2 = l2.next

            cur.val = sum % 10
            sum /= 10
            if l1 != None or l2 != None or sum != 0:
                cur.next = ListNode(0)
                cur = cur.next
            else:
                break

        return ret


# debug
s = Solution()

换算成整数求和

将两个链表直接换算成两个整数
这样的话，我们两个整数可以直接求和，而不用考虑各位的进位问题
最后再进行处理，将和再转换为链表

#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;

#define DEBUG



#ifdef DEBUG
class ListNode
{
public :
    ListNode( )
    :val(0), next(NULL)
    {
    }

    ListNode(int x)
    :val(x), next(NULL)
    {
    }

    int val;
    ListNode *next;


};

#endif




/** * 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)
 {  
        // IMPORTANT: Please reset any member data you declared, as 
        // the same Solution instance will be reused for each test case. 

        if((l1 == NULL) && (l2==NULL))
        {  
            return NULL;  
        }  

        unsigned long long sum1 = 0;  
        unsigned long long sum2 = 0;  
        unsigned long long sum3 = 0;  

        // 位数
        int count1 = 0; 
        int count2 = 0;  

        // 2 -> 4 -> 3
        while(l1 != NULL)  
        {  
            sum1 += pow(10, count1) * (l1->val);  
            count1++;  
            l1 = l1->next;  
        }  
        while(l2 != NULL)  
        {
            sum2 += pow(10, count2) * (l2->val);  
            count2++;  
            l2=l2->next;  
        }  

        sum3 = sum1 + sum2;  

#ifdef DEBUG
        std::cout <<sum1 <<" + " <<sum2 <<" = " <<sum3 <<std::endl;
#endif
        unsigned long tmp_digit = sum3 % 10;  
        sum3 = sum3 / 10;  

        ListNode* p = new ListNode((int)tmp_digit);  
        ListNode* start = p;  

        while(sum3 > 0)
        {  

            tmp_digit = sum3 % 10;
            sum3 /= 10;  

            p->next = new ListNode((int)tmp_digit);  
            p = p->next;  
        }  
        p->next = NULL;
        //printList(start); 
        return start;  
    }
};


#ifdef DEBUG

int main( int argc, char *argv[] )
{
    ListNode l1[3], l2[3];
    ListNode *pl1 = &l1[0], *pl2 = &l2[0];

    // Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
    // Output: 7 -> 0 -> 8
    l1[0].val = 2;l1[0].next = &l1[1];
    l1[1].val = 4;l1[1].next = &l1[2];
    l1[2].val = 3;l1[2].next = NULL;

    l2[0].val = 5;l2[0].next = &l2[1];
    l2[1].val = 6;l2[1].next = &l2[2];
    l2[2].val = 4;l2[2].next = NULL;

    Solution solu = Solution();
    ListNode *currNode = solu.addTwoNumbers(pl1, pl2);


    std::cout <<currNode->val;  
    while(currNode->next != NULL)   
    {
        currNode = currNode->next;
        std::cout <<" -> " <<currNode->val;
    }


    return EXIT_SUCCESS;
}
#endif