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[leetcode] 2. 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
Solution: link list
简单，注意两点：
1. 加到最后，可能还有进位
2. 链表实现时，可以多申请一个node的空间为x，返回x.next，可使代码更简洁。
#include <iostream>#include <vector>// 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) { ListNode out = ListNode(0); ListNode *tail = &out; if(!l1 || !l2) { return NULL; } int carry = 0; while(l1 && l2) { int s = carry + l1->val + l2->val; carry = s / 10; s = s % 10; l1 = l1->next; l2 = l2->next; ListNode *temp = new ListNode(s); tail->next = temp; tail = tail->next; } if(l2) { l1 = l2; } while(l1) { int s = l1->val + carry; carry = s / 10; s = s % 10; l1 = l1->next; ListNode *temp = new ListNode(s); tail->next = temp; tail = tail->next; } if(carry) { ListNode *temp = new ListNode(carry); tail->next = temp; tail = tail->next; } return out.next; }};
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