链表11

- 日理万妓 2022-05-25 03:07 78阅读 0赞

1、题目

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:  
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
    Output: 7 -> 8 -> 0 -> 7

2、解答: 刚开始还没看明白题目的意思；题目的大概意思是 两个数累加，你可以理解为 7243 + 564这中思想。思想是： 模型加法

3、C++代码：

class Solution {
    public:
        ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
            stack<int> stack1,stack2;
            while(l1 || l2){
                if(l1 != NULL){
                    stack1.push(l1->val);
                    l1 = l1->next;
                }
                if(l2 != NULL){
                    stack2.push(l2->val);
                    l2 = l2->next;
                }
            }
            
            int carry = 0;          //进位值
            ListNode *curr = NULL;
            
            while(!stack1.empty() || !stack2.empty()){
                int num1 = 0,num2 = 0;
                if(!stack1.empty()){
                    num1 = stack1.top();
                    stack1.pop();
                }
                if(!stack2.empty()){
                    num2 = stack2.top();
                    stack2.pop();
                }
                int sum = num1 + num2 + carry;
                //链表的头插法，以便更好的输出
                ListNode *temp = curr;
                curr = new ListNode(sum % 10);
                curr->next = temp;
                carry = sum / 10;
            }
            if(carry){
                ListNode *temp = curr;
                curr = new ListNode(carry);
                curr->next = temp;
            }
            return curr;
        }
    };

python代码  模型 加法的思想

def addTwoNumbers(self, l1, l2):
            """
            :type l1: ListNode
            :type l2: ListNode
            :rtype: ListNode
            """
            #该算法的思想是：模型加法
            x1,x2 = 0,0
            
            while l1:
                x1  = x1*10 + l1.val
                l1 = l1.next
            
            while l2:
                x2 = x2*10 + l2.val
                l2 = l2.next
            x = x1 + x2
            head = ListNode(0)
            if x  == 0:
                return head
            
            while x:
                v,x = x%10,x//10
                head.next,head.next.next = ListNode(v),head.next  #链表的头插法
            head = head.next
            return head