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2、Add Two Numbers

时间:2016-05-13 00:02:34      阅读:394      评论:0      收藏:0      [点我收藏+]

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1、Add Two Numbers——这是leedcode的第二题:
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

给定两个list,list里面的数字是按照逆序存储的,并且每一位包含一个数字。将两个list相加,返回一个list作为结果。这两个list相加的逻辑如下:各个位相加,逢十进一。

解决思路:
1、首先这是两个链表的相加问题。因为这里没有说链表谁长谁短,所以将链表先对应的位置相加赋值给结果链表,然后长的那个链表直接赋值给结果链表。
2、由于每个位置的数字只能是0-9的数字,所以会产生进位。所以链表在相加的时候,需要加上从前面得到的进位数。
3、如果最后加完所有的链表位置,还有进位制,这时候这个进位单独产生一个结点。

ps:其实就是两个数的加法运算,例子中的输入(逆序的)其实就是342+465的计算。
我们在计算的步骤分解:
2+5=7;
4+6=10;进一,得到0
2+5+1(进位)=8;
所以得到 807,逆序后就是708。

这里需要注意边界情况,例如:输入是[1]和[9,9]。

这里贴出Accept的代码:

Java版本

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode l3 = new ListNode(0);
        ListNode result = new ListNode(-1);
        int carry = 0;
        while (l1 != null && l2 != null) {
            int val = l1.val + l2.val + carry;
            carry = 0;
            if (val >= 10) {
                carry = 1;
            }
            ListNode node = new ListNode(val % 10);
            if (result.val == -1) {
                result.val = 0;
                result.next = node;
            }
            l3.next = node;
            l3 = node;
            l1 = l1.next;
            l2 = l2.next;
        }

        while (l1 != null) {
            int val = l1.val + carry;
            carry = 0;
            if (val >= 10) {
                carry = 1;
            }
            ListNode node = new ListNode(val % 10);
            l3.next = node;
            l3 = node;
            l1 = l1.next;
        }
        while (l2 != null) {
            int val = l2.val + carry;
            carry = 0;
            if (val >= 10) {
                carry = 1;
            }
            ListNode node = new ListNode(val % 10);
            l3.next = node;
            l3 = node;
            l2 = l2.next;
        }
        if (carry > 0) {
            ListNode node = new ListNode(carry % 10);
            l3.next = node;
            l3 = node;
        }
        result = result.next;
        return result;
    }
}

C++版本:

/**
 * 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* start=NULL;
        ListNode* l3 = NULL;
        int carry = 0;
        while (l1 != NULL&&l2 != NULL) 
        {
            int value = l1->val + l2->val + carry;
            carry = 0;
            if (value >= 10) {
                carry = 1;
            }
            ListNode* node = new ListNode(value % 10);
            if (start == NULL&&l3 == NULL) {
                    start = node;
                    l3 = node;
            }
            else
            {
                l3->next = node;
                l3 = l3->next;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1 != NULL) {
            int value = l1->val  + carry;
            carry = 0;
            if (value >= 10) {
                carry = 1;
            }
            ListNode* node = new ListNode(value % 10);
            if (start == NULL&&l3 == NULL) {
                start = node;
                l3 = node;
            }
            else
            {
                l3->next = node;
                l3 = l3->next;
            }
            l1 = l1->next;
        }
        while (l2 != NULL) {
            int value = l2->val + carry;
            carry = 0;
            if (value >= 10) {
                carry = 1;
            }
            ListNode* node = new ListNode(value % 10);
            if (start == NULL&&l3 == NULL) {
                start = node;
                l3 = node;
            }
            else
            {
                l3->next = node;
                l3 = l3->next;
            }
            l2 = l2->next;
        }
        if (carry > 0) 
        {
            ListNode* node = new ListNode(carry % 10);
            carry = 0;
            if (start == NULL&&l3 == NULL) {
                start = node;
                l3 = node;
            }
            else
            {
                l3->next = node;
                l3 = l3->next;
            }
        }
        return start;
    }
};

2、Add Two Numbers

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原文地址:http://blog.csdn.net/buptzhengchaojie/article/details/51347930

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