［LeetCode］Add Two Numbers

解題思路： 1，建立一個preHead，這樣方便處理第一次運算； 2，加一個carry用於儲存 進位元； 3，每一位的運算，都包括 l1.val + l2.val + carry 4，每一次運算結果，都通過 除法和求餘運算，得到 carry和當前位的數字

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param {ListNode} l1    # @param {ListNode} l2    # @return {ListNode}    def addTwoNumbers(self, l1, l2):        dummy = temp = ListNode(0)        carry = 0        while l1 or l2 or carry:            res = 0            if l1:                res += l1.val                l1 = l1.next            if l2:                res += l2.val                l2 = l2.next            res += carry            temp.next = ListNode(res % 10)            temp = temp.next            carry = res // 10        return dummy.next

// Java 實現

public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode dummy = new ListNode(0);        ListNode p = dummy;        int carry = 0;        while(l1 != null || l2 != null){            int x = (l1 != null ? l1.val : 0);            int y = (l2 != null ? l2.val : 0);            int ret = x + y + carry;            p.next = new ListNode(ret%10);            p = p.next;            carry = ret/10;            if (l1 != null) l1 = l1.next;            if (l2 != null) l2 = l2.next;        }        if (carry > 0){            p.next = new ListNode(carry);        }        return dummy.next;    }}