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 <val;
while(currNode->next != NULL)
{
currNode = currNode->next;
std::cout <<" -> " <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 " + " <" = " <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 <val;
while(currNode->next != NULL)
{
currNode = currNode->next;
std::cout <<" -> " <val;
}
return EXIT_SUCCESS;
}
#endif