2. Add Two Numbers

屏幕快照 2022-04-09 下午8.57.35.png

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;
    }
}