题目描述

解题思路

Java代码

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode res = null; //结果链表，不可移动，一旦移动就会丢失值
        ListNode cur = null; //结果链表的游标，可移动，为了往结果链表中添加
        int upval = 0; //进位值
        //情况一
        while(l1 != null && l2 != null){
            int val1 = l1.val;
            int val2 = l2.val;
            int val = val1 + val2 + upval;
            
            upval = val / 10;
            int curval = val % 10;
            
            ListNode thisNode = new ListNode(curval);
            
            if(res == null){
                res = thisNode;
                cur = res;
            }else{
                cur.next = thisNode;
                cur = thisNode;
            }
            l1 = l1.next;
            l2 = l2.next;
        }
        //情况二
        while(l1 == null && l2 != null){  //l1长度小于l2
            int val = l2.val + upval;
            upval = val / 10;
            int curval = val % 10;
            
            ListNode thisNode = new ListNode(curval);
            
            if(res == null){
                res = thisNode;
                cur = res;
            }else{
                cur.next = thisNode;
                cur = thisNode;
            }
            l2 = l2.next;
            
        }
        //情况三
        while(l2 == null && l1 != null){  //l2长度小于l1
            int val = l1.val + upval;
            upval = val / 10;
            int curval = val % 10;
            
            ListNode thisNode = new ListNode(curval);
            
            if(res == null){
                res = thisNode;
                cur = res;
            }else{
                cur.next = thisNode;
                cur = thisNode;
            }
            l1 = l1.next;
        }
        
        if(upval != 0){
            ListNode thisNode = new ListNode(upval);
            cur.next = thisNode;
        }
        
        return res;
    }
}

此代码还可以整合下，我这么写主要是为了看起来和图示搭配，可以按下面两个代码的方式写，看起来简单点

Python3代码

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 is None:
            return l2
        if l2 is None:
            return l1
        
        res = ListNode(0)
        cur = res
        upval = 0
        while l1 or l2:
            tmpsum = 0
            if l1:  # 情况二
                tmpsum = l1.val
                l1 = l1.next
            if l2:  # 情况三
                tmpsum += l2.val
                l2 = l2.next
            curval = (tmpsum + upval) % 10
            upval = (tmpsum + upval) // 10
            cur.next = ListNode(curval)
            cur = cur.next
        # 最后一位是否进位
        if upval:
            cur.next = ListNode(1)
        del cur
        res = res.next
        return res

C++代码

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if(l1 == NULL){
            return l2;
        }
        if(l2 == NULL){
            return l1;
        }
        
        ListNode* res = new ListNode(0);
        ListNode* cur = res;
        int upval = 0;
        
        while(l1 != NULL || l2 != NULL){
            int tmpsum = 0;
            if(l1 != NULL){
                tmpsum = l1->val;
                l1 = l1->next;
            }
            if(l2 != NULL){
                tmpsum += l2->val;
                l2 = l2->next;
            }
            int curval = (tmpsum + upval) % 10;
            upval = (tmpsum + upval) / 10;
            cur->next = new ListNode(curval);
            cur = cur->next;
        }
        if(upval != 0){
            cur->next = new ListNode(1);
        }
        res = res->next;
        return res;
    }
};