leetcode-002 Add Two Numbers

P002-Add-Two-Numbers
思路分析

代码
java

python

P002-Add-Two-Numbers

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

思路分析

就是两个链表的并行遍历，并不怎么难。

肯定得遍历两个链表

两链表长度相同时，对应位置相加并将进位保留到下一次循环

两链表长度不相同时，往长度较短的链表尾部补零(最高位和现实生活中的数字恰好相反)

代码

java

/***
* 2 4 3        2 4 3 0     1 2 3 4 0
* 5 6 4        5 6 4 9     6 7 8 9 9
* -------      -------     ----------
* 7 0 8        7 0 8 9     7 9 1 4 0 1
*
* **/
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null)
return l2;
if (l2 == null)
return l1;
ListNode l = l1;
ListNode r = l2;
ListNode ret = new ListNode(0);
int up = 0;
int currentVal = 0;
ListNode last = ret;
while (l != null || r != null) {

// 两个链表长度不一致：补零
if (l == null)
l = new ListNode(0);
if (r == null)
r = new ListNode(0);

// 当前值:加上进位
currentVal = (l.val + r.val) + up;

if (currentVal >= 10) {
up = currentVal / 10;// 为下一轮提供的进位
currentVal = currentVal % 10;// 当前值
} else {
up = 0;
}

// 新节点
ListNode newNode = new ListNode(currentVal);

last.next = newNode;
last = last.next;

l = l.next;
r = r.next;
}
if (up != 0) {
last.next = new ListNode(up);
}
return ret.next;
}

python

def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if (not l1) : return l2
if (not l2) : return l1
l = l1;r = l2;
ret = ListNode(0);last = ret
up = 0;currentVal = 0

while (l or r):
if not l:l = ListNode(0)
if not r:r = ListNode(0)

currentVal = (l.val + r.val) + up

if currentVal >= 10:
up = currentVal / 10
currentVal = currentVal % 10
else:
up = 0

newNode = ListNode(currentVal)
last.next = newNode
last = last.next

l = l.next
r = r.next

# end while

if up != 0:
last.next = ListNode(up)

return ret.next