2. 两数相加 @Python @Go

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(1)

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

Go代码

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	dummy := ListNode{}
    // 当前指针
	cur := &dummy
	carry := 0
	for l1 != nil || l2 != nil || carry != 0 {
		total := carry
		if l1 != nil {
			total += l1.Val
			l1 = l1.Next
		}
		if l2 != nil {
			total += l2.Val
			l2 = l2.Next
		}
		cur.Next = &ListNode{Val: total % 10}
		// 进位
		carry = total / 10
		// 移动指针
		cur = cur.Next
	}
	return dummy.Next
}