leetcode_445

Golang:

思路：组合题，先反转链表，再让它们相加，再反转回来即可。针对题目提到的进阶，可以先将两个链表拷贝下来做大数加法，再赋值回去或者新开一条链表

代码如下：

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    l1 = reverseList(l1)
    l2 = reverseList(l2)
    temp:=addTwoLinkedLists(l1,l2)
    return reverseList(temp)
}

//两个链表相加，使用的某大佬的代码
func addTwoLinkedLists(l1 *ListNode, l2 *ListNode) *ListNode {
    var temp = &ListNode{}
    var incr int

    lastNode := temp
    for {
        var v int
        if l1 != nil && l2 != nil {
            v = l1.Val + l2.Val + incr
            if v > 9 {
                v = v - 10
                incr = 1
            } else {
                incr = 0
            }
            l1 = l1.Next
            l2 = l2.Next
        } else if l1 != nil {
            v = l1.Val + incr
            if v > 9 {
                v = v - 10
                incr = 1
            } else {
                incr = 0
            }
            l1 = l1.Next
        } else if l2 != nil {
            v = l2.Val + incr
            if v > 9 {
                v = v - 10
                incr = 1
            } else {
                incr = 0
            }
            l2 = l2.Next
        } else {
            break
        }
        node := &ListNode{
            Val: v,
        }
        lastNode.Next = node

        lastNode = node
    }

    if incr == 1 {
        node := &ListNode{
            Val: 1,
        }
        lastNode.Next = node
    }

    return temp.Next

}

//翻转链表，用的某位大佬的代码
func reverseList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }

    var pre *ListNode = nil
    var cur *ListNode = head
    for cur != nil {
        tmp := cur.Next
        cur.Next = pre
        pre = cur
        cur = tmp
    }
    return pre
}

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