2. 两数相加 @python @java

原题

https://leetcode.cn/problems/add-two-numbers/

思路

模拟法，从左向右相加，如果和超过10就进位。

复杂度

时间：O(n)
空间：O(n)

Python代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        pre = ListNode(0)
        dummy = pre
        carry = 0
        while l1 or l2:
            n1 = l1.val if l1 is not None else 0
            n2 = l2.val if l2 is not None else 0
            # 当前位数字
            val = (n1 + n2 + carry) % 10
            # 进位数字
            carry = (n1 + n2 + carry) // 10
            # 生成节点
            pre.next = ListNode(val)
            # 后移一位
            pre = pre.next
            if l1:
                l1 = l1.next 
            if l2: 
                l2 = l2.next 
        # 如果carry不为0，加上carry
        if carry > 0:
            pre.next = ListNode(carry)
        return dummy.next

Java代码

/**
 * 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) {
        // 生成虚拟节点
        ListNode pre = new ListNode(0);
        ListNode dummy = pre;
        // 进位数字
        int carry = 0;
        while (l1 != null || l2 != null) {
            int n1 = l1 != null? l1.val : 0;
            int n2 = l2 != null? l2.val : 0;
            int val = (n1 + n2 + carry) % 10;
            carry = (n1 + n2 + carry)/ 10;
            // 生成节点
            pre.next = new ListNode(val);
            // 移动指针
            pre = pre.next;
            if (l1 != null)            
                l1 = l1.next; 
            if (l2 != null)           
                l2 = l2.next;
        }    
        if (carry > 0) 
            pre.next = new ListNode(carry);
        return dummy.next;
    }
}