2022-02-24 025，024 链表反转,链表中的两数相

链表反转：
思路采用栈，而栈可以用slice的思路实现
Go版本

func reverseList(head *ListNode) *ListNode {

    // 数组slice当作stack
    stack:=[]int{}
    for head!=nil{
        stack=append(stack,head.Val);
        head=head.Next
    }
    if len(stack)==0{
        return nil
    }
    ans:=&ListNode{Val:stack[len(stack)-1]};
    res:=ans;
    stack=stack[:len(stack)-1]
    for len(stack)>0{
        temp:=stack[len(stack)-1]
        stack=stack[:len(stack)-1]
        node:=&ListNode{Val:temp}
        res.Next=node
        res=res.Next
    }
    return ans;

}

链表两数相加：
纯粹的反转相加，注意考虑类似于l1=[5],l2=[5]的情况
Go版本

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {

    // l1,l2非空链表反转
    l1_reverse,l1_count:=reverse(l1);
    l2_reverse,l2_count:=reverse(l2);
    ans:=&ListNode{Val:0}
    res:=ans;
    min_count:=0;
    // 计算链表长度
    if l1_count>l2_count{
        min_count=l2_count
        // max_count=l1_count;
    }else{
        min_count=l1_count
        // max_count=l2_count-l1_count;
    }
    temp:=0 //表示进

    for min_count>0{
        Value:=l1_reverse.Val+l2_reverse.Val+temp
        temp=0;
        if Value>=10{
           Value=Value-10;
           temp=1;  
        }else{
            Value=Value;
        }
        //  fmt.Print(l1_reverse.Val,l2_reverse.Val,Value)
        //  fmt.Println()
         
        node:=&ListNode{Val:Value}
        l1_reverse=l1_reverse.Next
        l2_reverse=l2_reverse.Next
        res.Next=node;
        res=res.Next;
        min_count--;
        l1_count--;
        l2_count--;   
    }
  
    LastNode:=&ListNode{}
    if l1_count!=0{
        LastNode=l1_reverse;
    }else{
        LastNode=l2_reverse;
    }
   
    for LastNode!=nil || temp!=0{
         if LastNode==nil{
                LastNode=&ListNode{Val:0};
                
            }
        
         Value:=LastNode.Val+temp
          temp=0;
        if Value>=10{
           Value=Value-10;
           temp=1;  
        }else{
            Value=Value;
        }
        node:=&ListNode{Val:Value}
        res.Next=node;
        res=res.Next;
        LastNode=LastNode.Next
        
    }
    ans_reverse,_:=reverse(ans.Next)
return ans_reverse;

}
func reverse(head *ListNode)(*ListNode,int){
    stack:=[]int{};
    count:=0
    for head!=nil{
        stack=append(stack,head.Val)
        head=head.Next
        count++;
    }
    ans:=&ListNode{Val:stack[len(stack)-1]}
    stack=stack[:len(stack)-1]
    res:=ans;
    for(len(stack)>0){
        node:=&ListNode{Val:stack[len(stack)-1]}
        stack=stack[:len(stack)-1]
        res.Next=node
        res=res.Next
        
    }
    return ans,count;
}