lintcode 访问地址
http://www.lintcode.com/zh-cn/problem/add-two-numbers/
描述
你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。
样例
给出两个链表3->1->5->null和5->9->2->null，返回8->0->8->null。
Java代码实现
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| /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
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
if (l1 == null && l2 == null) {
return null;
}
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 = carry / 10;
if (l1 != null || l2 != null || carry != 0) {
ptr.next = new ListNode(0);
ptr = ptr.next;
} else {
break;
}
}
return head;
}
}
|
Python代码实现
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| # Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param l1: the first list
# @param l2: the second list
# @return: the sum list of l1 and l2
def addLists(self, l1, l2):
# write your code here
head = ListNode(0)
ptr = head
carry = 0
while True:
if l1 is not None:
carry += l1.val
l1 = l1.next
if l2 is not None:
carry += l2.val
l2 = l2.next
ptr.val = carry % 10
carry = carry / 10
if l1 is not None or l2 is not None or carry != 0:
ptr.next = ListNode(0)
ptr = ptr.next
else:
break
return head
|