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题目来源：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
题目要求：给出两个链表l1和l2,将相对应的元素相加,如果产生进位则保留个位,相应的十位加在后面的对应元素上.
解体思路：
首先判断链表是否为空,若l1空,直接返回l2,相反返回l2
然后找到l1和l2链表最短的一个长度min_len,进入循环
循环过程中,为了减少空间的开销,可以将链表l1作为最终的结果链表,只需要改变l1的最大容量为l1+l2即可,另外相加过程中需要一个整型temp变量来保存进位.
循环结束之后,判断l1或l2空否(之前计算最短链表时可以做标记),然后将不空的链表中其他元素全部放入结果链表中即可.
提交代码：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 10 class Solution { 11 public: 12 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { 13 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 // ListNode *pResult = NULL; 16 // ListNode **pCur = &pResult; 17 18 ListNode rootNode(0); 19 ListNode *pCurNode = &rootNode; 20 int a = 0; 21 while (l1 || l2) 22 { 23 int v1 = (l1 ? l1->val : 0); 24 int v2 = (l2 ? l2->val : 0); 25 int temp = v1 + v2 + a; 26 a = temp / 10; 27 ListNode *pNode = new ListNode((temp % 10)); 28 pCurNode->next = pNode; 29 pCurNode = pNode; 30 if (l1) 31 l1 = l1->next; 32 if (l2) 33 l2 = l2->next; 34 } 35 if (a > 0) 36 { 37 ListNode *pNode = new ListNode(a); 38 pCurNode->next = pNode; 39 } 40 return rootNode.next; 41 } 42 };
下面给出数组解决的代码,具体过程相同,只是数据存储结构不同：
1 #include <bits/stdc++.h> 2 #define MAX 1000010 3 4 using namespace std; 5 6 int main() 7 { 8 int n1,n2; 9 while(~scanf("%d %d",&n1,&n2)) 10 { 11 int *a=new int[n1+n2+1]; 12 int *b=new int[n2]; 13 for(int i=0;i<n1;i++) 14 scanf("%d",&a[i]); 15 for(int i=0;i<n2;i++) 16 scanf("%d",&b[i]); 17 int minlen=min(n1,n2); 18 int maxlen=n1+n2-minlen; 19 int temp=0; 20 for(int i=0;i<minlen;i++) 21 { 22 a[i]=a[i]+b[i]+temp; 23 if(a[i]>=10) 24 { 25 temp=a[i]/10; 26 a[i]=a[i]%10; 27 } 28 } 29 for(int j=minlen;j<maxlen;j++) 30 { 31 if(maxlen==n1) 32 a[j]=a[j]; 33 else 34 a[j]=b[j]; 35 } 36 for(int i=0;i<maxlen;i++) 37 printf("%d ",a[i]); 38 printf("\n"); 39 } 40 }
LeetCode 2 Add Two Numbers（链表操作）
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原文地址：http://www.cnblogs.com/zpfbuaa/p/5049623.html
