[LeetCode] Add Two Numbers（链表合并+模拟加法）

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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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

题意：给出两个链表，每一个链表代表着一个数，要求把两个链表所得到的数相加，得到第三个链表，注意链表的第一个数为这个数的最后一位，没有前导零，除了0本身。

思路：一开始想着把两个链表所表示的数算出来再相加，再转化为链表，然后发现样例中有很大很大的数，只能模拟加法，模拟加法的时候注意进位。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
	public ListNode addTwoNumbers(ListNode l1,ListNode l2) {
                //设置一个空节点
		ListNode l3 = new ListNode(0),head = l3;
		int add = 0,temp;
		while(l1 != null && l2 != null) {
			if(add + l1.val + l2.val >= 10) {
                                temp = add + l1.val + l2.val - 10;
				add = 1;
			}
			else {
                                temp = add + l1.val + l2.val;
				add = 0;
			}
			l3.next = new ListNode(temp);
			l3 = l3.next;
			l1 = l1.next;
			l2 = l2.next;
		}
		while(l1 != null) {
			if(add + l1.val >= 10) {
                                temp = add + l1.val - 10;
				add = 1;
			}
			else {
                                temp = add + l1.val;
				add = 0;
			}
			l3.next = new ListNode(temp);
			l3 = l3.next;
			l1 = l1.next;
		}
		while(l2 != null) {
			if(add + l2.val >= 10) {
                                temp = add + l2.val - 10;
				add = 1;
			}
			else {
                                temp = add + l2.val;
				add = 0;
			}
			l3.next = new ListNode(temp);
			l3 = l3.next;
			l2 = l2.next;
		}
		if(add == 1) {
			l3.next = new ListNode(1);
			l3 = l3.next;
			add = 0;
		}
		l3.next = null;
		return head.next;
	}
}

C语言代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    //ListNode l3 = new ListNode(0),head = l3;
        struct ListNode *l3, *head;
        l3 = (struct ListNode*)malloc(sizeof(struct ListNode));
        head = l3;
	int add = 0,temp;
	while(l1 != NULL && l2 != NULL) {
	    if(add + l1 -> val + l2 -> val >= 10) {
                temp = add + l1 -> val + l2 -> val - 10;
		add = 1;
	    }
	    else {
                temp = add + l1 -> val + l2 -> val;
		add = 0;
	    }
            l3 -> next = (struct ListNode*) malloc(sizeof(struct ListNode));
            l3 -> next -> val = temp;
	    l3 = l3 -> next;
	    l1 = l1 -> next;
	    l2 = l2 -> next;
	}
	while(l1 != NULL) {
		if(add + l1 -> val >= 10) {
                    temp = add + l1 -> val - 10;
		    add = 1;
		}
		else {
                    temp = add + l1 -> val;
		    add = 0;
		}
		l3 -> next = (struct ListNode*) malloc(sizeof(struct ListNode));
                l3 -> next -> val = temp;
		l3  = l3 -> next;
		l1 = l1 -> next;
	}
	while(l2 != NULL) {
		if(add + l2 -> val >= 10) {
                    temp = add + l2 -> val - 10;
		    add = 1;
		}
		else {
                    temp = add + l2 -> val;
		    add = 0;
		}
		l3 -> next = (struct ListNode*) malloc(sizeof(struct ListNode));
                l3 -> next -> val = temp;
		l3 = l3 -> next;
		l2 = l2 -> next;
	}
	if(add == 1) {
		l3 -> next = (struct ListNode*) malloc(sizeof(struct ListNode));
                l3 -> next -> val = 1;
		l3 = l3 -> next;
		add = 0;
	}
	l3 -> next = NULL;
	return head -> next;
}