标签：des style blog io color ar 使用 for sp
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：两条链表相加，注意最后如果进位不为零，要加入到结果中。使用二级指针记录节点前缀。
/** * 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 || l2 == NULL) { return l1 == NULL ? l1 : l2; } int carry = 0; ListNode *head = NULL; ListNode **prev = &head; while (l1 != NULL || l2 != NULL) { int sum = carry; int l1_val = 0; if(l1 != NULL) { l1_val = l1->val; l1 = l1->next; } int l2_val = 0; if (l2 != NULL) { l2_val = l2->val; l2 = l2->next; } sum += l1_val + l2_val; ListNode *pt = new ListNode(sum % 10); *prev = pt; prev = &(pt->next); carry = sum / 10; } if (carry) { ListNode *pt = new ListNode(carry); *prev = pt; } return head; } };
《LeetCode 题解 （C++版）》（戴方勤 著 ）中一种非常简洁的写法：
1 class Solution { 2 public: 3 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { 4 ListNode head(-1); // 头节点 5 int carry = 0; 6 ListNode *prev = &head; 7 for (ListNode *pa = l1, *pb = l2; 8 pa != nullptr || pb != nullptr; 9 pa = pa == nullptr ? nullptr : pa->next, 10 pb = pb == nullptr ? nullptr : pb->next, 11 prev = prev->next) { 12 const int ai = pa == nullptr ? 0 : pa->val; 13 const int bi = pb == nullptr ? 0 : pb->val; 14 const int value = (ai + bi + carry) % 10; 15 carry = (ai + bi + carry) / 10; 16 prev->next = new ListNode(value); // 尾插法 17 } 18 if (carry > 0) 19 prev->next = new ListNode(carry); 20 21 return head.next; 22 } 23 }；
标签：des style blog io color ar 使用 for sp
原文地址：http://www.cnblogs.com/vincently/p/4055340.html
