LeetCode-2： Add Two Numbers

【Problem：2-Add Two Numbers】

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

【Solution】

1）From 九章算法：（can run, but "Submission Result：  Wrong Answer！ "）

class Solution:
    def addTwoNumbers(self, l1, l2):
        head = ListNode(0)
        ptr = head
        carry  = 0
        while True:
            if l1 != None:
                carry += l1.val
                l1 = l1.next
            if l2 != None:
                carry += l2.val
                l2 = l2.next
            ptr.val = carry % 10
            carry /= 10
            # 运算未结束新建一个节点用于储存答案，否则退出循环
            if l1 != None or l2 != None or carry != 0:
                ptr.next = ListNode(0)
                ptr = ptr.next
            else: 
                break
        return head

 2）Java ：

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


// version: 高频题班
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        // write your code here
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;

        int carry = 0;
        for (ListNode i = l1, j = l2; i != null || j != null; ) {
            int sum = carry;
            sum += (i != null) ? i.val : 0;
            sum += (j != null) ? j.val : 0;

            tail.next = new ListNode(sum % 10);
            tail = tail.next;

            carry = sum / 10;
            i = (i == null) ? i : i.next;
            j = (j == null) ? j : j.next;
        }

        if (carry != 0) {
            tail.next = new ListNode(carry);
        }
        return dummy.next;
    }
}

3）C++：yes, accepted

/**
 * 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) {
        // 题意可以认为是实现高精度加法
        ListNode *head = new ListNode(0);
        ListNode *ptr = head;
        int carry = 0;
        while (true) {
            if (l1 != NULL) {
                carry += l1->val;
                l1 = l1->next;
            }
            if (l2 != NULL) {
                carry += l2->val;
                l2 = l2->next;
            }
            ptr->val = carry % 10;
            carry /= 10;
            // 当两个表非空或者仍有进位时需要继续运算，否则退出循环
            if (l1 != NULL || l2 != NULL || carry != 0) {
                ptr = (ptr->next = new ListNode(0));
            } else break;
        }
        return head;
    }
};

【References】

1、http://www.jiuzhang.com/solution/add-two-numbers/

Time complexity：O(n^2)2