收获：1、程序调试一定要根据结果推理，不能乱试，至少要有根据的试
2、单向链表遍历的时候要注意: 不能找到之前的元素。这样，在遍历到list末尾的时候，就不能给list添加元素了。
Problem: 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
思路：链表对应的位置相加即可。
难点：1、加法的进位处理；
2、两个链表的遍历过程中在末尾如何处理。
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 ListNode* initList(); 10 void listAdd(ListNode* last, int val); 11 class Solution { 12 public: 13 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { 14 // IMPORTANT: Please reset any member data you declared, as 15 // the same Solution instance will be reused for each test case. 16 ListNode *L1 = l1; 17 ListNode *L2 = l2; 18 ListNode *temp = NULL; // to keep the L1 finding the list's end 19 int increase = 0; 20 while(L1 || L2 || increase){ 21 if(L1 && L2){ // L1 and L2 are not empty 22 int sum_val = L1->val + L2->val + increase; 23 L1->val = sum_val % 10; 24 increase = sum_val / 10; 25 temp = L1; // in case of L1 is the end of list 26 L1 = L1->next; 27 L2 = L2->next; 28 }else if((NULL == L1) && (NULL != L2)){ // L1 empty 29 L1 = new ListNode(0); 30 temp->next = L1; 31 int sum_val = L2->val + increase; 32 L1->val = sum_val % 10; 33 increase = sum_val/10; 34 temp = L1; 35 L1 = L1->next; 36 L2 = L2->next; 37 } 38 else if((NULL != L1) && (NULL == L2)){ 39 int sum_val = L1->val + increase; 40 L1->val = sum_val % 10; 41 increase = sum_val /10; 42 temp = L1; 43 L1 = L1->next; 44 } 45 else if(increase){ 46 L1 = new ListNode(1); 47 temp->next = L1; 48 //L1 = temp; 49 //L1->val = 1; 50 increase = 0; 51 L1 = L1->next; 52 } 53 } 54 return l1; 55 } 56 };
这个程序破坏了输入链表。这不好（根据Code Complete2）。下面的程序应该也可以。
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 ListNode* initList(); 10 void listAdd(ListNode* last, int val); 11 class Solution { 12 public: 13 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { 14 // IMPORTANT: Please reset any member data you declared, as 15 // the same Solution instance will be reused for each test case. 16 if((!l1) && (!l2)) return NULL; 17 ListNode *L1 = l1; 18 ListNode *L2 = l2; 19 ListNode *res = new ListNode(0); 20 ListNode *last = res; 21 ListNode *temp = res; 22 int flag = 0; 23 24 while(L1 || L2 || flag){ 25 if(L1 && L2){ // L1 and L2 are not empty 26 int sum_val = L1->val + L2->val + flag; 27 if(NULL == last){ 28 last = new ListNode(sum_val % 10); 29 temp->next = last; 30 temp = last; 31 }else{ 32 last->val = (sum_val % 10); // 33 } 34 flag = sum_val/10; 35 last = last->next; 36 L1 = L1->next; 37 L2 = L2->next; 38 }else if((NULL == L1) && (NULL != L2)){ 39 int sum_val = L2->val + flag; 40 if(NULL == last){ 41 last = new ListNode(sum_val % 10); 42 temp->next = last; 43 temp = last; 44 }else{ 45 last->val = (sum_val % 10); 46 } 47 flag = sum_val/10; 48 last = last->next; 49 L2 = L2->next; 50 } 51 else if((NULL != L1) && (NULL == L2)){ 52 L2 = L1; 53 L1 = NULL; 54 } 55 else if(flag){ 56 last = new ListNode(1); 57 temp->next = last; 58 flag = 0; 59 } 60 } 61 return res; 62 } 63 };