445. 两数相加 II(Medium)

Python示例

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        stack1, stack2 = [], []
        while l1:
            stack1.append(l1.val)
            l1 = l1.next 
        while l2:
            stack2.append(l2.val)
            l2 = l2.next 
        dummpyHead = ListNode(None)
        digit = 0
        while stack1 or stack2:
            a = stack1.pop() if stack1 else 0
            b = stack2.pop() if stack2 else 0
            s = a + b + digit 
            node = ListNode(s % 10)
            digit = s // 10 
            node.next = dummpyHead.next 
            dummpyHead.next = node 
        if digit > 0:
            node = ListNode(digit)
            node.next = dummpyHead.next 
            dummpyHead.next = node 
        return dummpyHead.next 

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    var stack1, stack2 []int
    for l1 != nil {
        stack1 = append(stack1, l1.Val)
        l1 = l1.Next
    }
    for l2 != nil {
        stack2 = append(stack2, l2.Val)
        l2 = l2.Next
    }
    
    var cur *ListNode
    var a, b int 
    carry := 0 
    aLen, bLen := len(stack1), len(stack2)
    for aLen != 0 || bLen != 0 {
        if aLen > 0 {
            a = stack1[aLen - 1]
            aLen--
        } else {
            a = 0
        }

        if bLen > 0 {
            b = stack2[bLen - 1]
            bLen--
        } else {
            b = 0
        }

        node := &ListNode{Val: (a + b + carry) % 10}
        carry = (a + b + carry) / 10
        node.Next = cur 
        cur = node 
    }

    if carry != 0 {
        node := &ListNode{Val: carry}
        node.Next = cur 
        cur = node
    }
    return cur 
}