https://leetcode.com/problems/add-two-numbers/
空间复杂度O(max(m,n))
时间复杂度O(max(m,n))
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
result = ListNode(0)
result_tail = result
carry = 0
while l1 or l2 or carry:
val1 = l1.val if l1 else 0
val2 = l2.val if l2 else 0
out = (val1 + val2 + carry) % 10
carry = (val1 + val2 + carry) // 10
result_tail.next = ListNode(out)
result_tail = result_tail.next
l1 = l1.next if l1 else None
l2 = l2.next if l2 else None
return result.next
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode result = new ListNode(0);
ListNode result_tail = result;
int carry = 0;
while (l1 != null || l2 != null || carry>0){
int l1_val = (l1 != null) ? l1.val : 0;
int l2_val = (l2 != null) ? l2.val : 0;
int out = (l1_val + l2_val + carry) % 10;
carry = (l1_val + l2_val + carry) / 10;
result_tail.next = new ListNode(out);
result_tail = result_tail.next;
l1 = (l1 != null) ? l1.next : null;
l2 = (l2 != null) ? l2.next : null;
}
return result.next;
}
}