[LeetCode 解题报告]002.Add Two Numbers

Description:

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Examples:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

算法思想：类似于大数加法，若当前和大于10，则产生进位。一直遍历到L1和L2尾部。注意：L1和L2的长度可能不一样长。最后也可能产生进位。

/** * 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* res= new ListNode(0);        ListNode* p = res;        if(l1 == NULL || l2 == NULL)            return NULL;        int c = 0;        while(l1 != NULL && l2 != NULL) {            p->next = new ListNode((c + l1->val + l2->val) %10);            c = (c + l1->val + l2->val) / 10;                        p = p->next;            l1 = l1->next;            l2 = l2->next;        }                while(l1 != NULL) {            p->next = new ListNode((c + l1->val) % 10);            c = (c + l1->val) / 10;            p = p->next;            l1 = l1->next;        }                while(l2 != NULL) {            p->next = new ListNode((c + l2->val) % 10);            c = (c + l2->val) / 10;            p = p->next;            l2 = l2->next;        }                if(c)            p->next = new ListNode(c);        return res->next;     }};

时间复杂度：O(max(m, n))，空间复杂度O(max(m, n) + 1)