Add Two Numbers 分类： Leetcode（链表） 2015-02-17 11:09 43人阅读 评论(0) 收藏

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

很简单的一道题，最直观的就是一一相加，然后再把多于的链表连接上，注意特殊的情况比如{5} {5},链表的最后得加上进位1

/**
 * 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* pre = new ListNode(-1);
        ListNode* head = pre;
        int carry = 0;
        while (l1 || l2){
            int val1 ;
            int val2 ;
            if(l1){
                val1 = l1->val;
                l1 = l1->next;
            }
            else
                val1 = 0;
            if(l2){
                val2 = l2->val;
                l2 = l2->next;
            }
            else val2 = 0;
            ListNode* node = new ListNode( (val1 + val2 + carry)%10);
            pre->next = node;
            pre = pre->next;
            carry = (val1 + val2 + carry) / 10;
        }
        if(carry){
            ListNode* node = new ListNode(carry);
            pre->next = node;
        }
        return head->next;
  
    }
};

/**
 * 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* pre = NULL;
        ListNode* head = NULL;
        int carry = 0;
        while (l1 || l2){
            int val1 ;
            int val2 ;
            if(l1){
                val1 = l1->val;
                l1 = l1->next;
            }
            else
                val1 = 0;
            if(l2){
                val2 = l2->val;
                l2 = l2->next;
            }
            else val2 = 0;
            ListNode* node = new ListNode( (val1 + val2 + carry)%10);
            if(head == NULL){
                head = node;
            }
            else{
                pre->next = node;
            }
            pre = node;
        
            carry = (val1 + val2 + carry) / 10;
        }
        if(carry){
            ListNode* node = new ListNode(carry);
            pre->next = node;
        }
        return head;
  
    }
};

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