1 题目

2 思路

遍历链表，逐位进行累加，每次记录进位数，加入到下一位的累加中，模拟加法运算，边累加边使用尾插法，进行结果链表的创建，最后返回结果链表。

3 代码

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        //ListNode* ans = (ListNode*) malloc(sizeof(ListNode));
        ListNode* ans = new ListNode;
        ListNode* r = ans;
        bool isFisrtNode = true;
        int carry = 0;//进位标识符
        while(l1 != nullptr && l2 != nullptr){
            int tmpValue = carry;
            carry = 0;
            tmpValue += l1->val + l2->val;
            if(tmpValue > 9){
                carry = tmpValue/10;
                tmpValue %= 10;
            }else{}
            if(!isFisrtNode){
                ListNode* s = new ListNode;
                s->val = tmpValue;
                r->next = s;
                r = s;
            }else{
                r->val = tmpValue;
                isFisrtNode = false;
            }
            l1 = l1->next;
            l2 = l2->next;
        }

        //处理多余的位数（注意处理进位的结果）
        while(l1 != nullptr){
            ListNode* s =  new ListNode;
            int tmpValue = l1->val;
            if(carry != 0){
                tmpValue += carry;
                carry = 0;
                if(tmpValue > 9){
                    carry = tmpValue/10;
                    tmpValue %= 10;
                }else{}
            }
            s->val = tmpValue;
            r->next = s;
            r = s;
            l1 = l1->next;
        }

        while(l2 != nullptr){
            ListNode* s = new ListNode;

            int tmpValue =  l2->val;
            if(carry != 0){
                tmpValue += carry;
                carry = 0;
                if(tmpValue > 9){
                    carry = tmpValue/10;
                    tmpValue %= 10;
                }else{}
            }

            s->val = tmpValue;
            r->next = s;
            r = s;
            l2 = l2->next;
        }
        
        //处理最后进位的结果
        if(carry != 0){
            ListNode* s = new ListNode;
            s->val = carry;
            r->next = s;
            r = s;
        }

        r->next = nullptr;

        return ans;
    }
};

4 实例

#include <iostream>
#include <vector>
#include <deque>
using namespace std;
//数据结构
struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

//尾插法创建链表
ListNode* createListNode(deque<int> input){
    ListNode* p = (ListNode*) malloc(sizeof(ListNode));
    ListNode* r = p;
    p->val = input.front();
    input.pop_front();
    while(!input.empty()){
        ListNode* s = (ListNode*) malloc(sizeof(ListNode));
        s->val = input.front();
        input.pop_front();
        r->next = s;
        r = s;
    }
    r->next = nullptr;
    return p;
}

//打印链表
void printListNode(ListNode* p){
    ListNode* q = p;
    while(q != nullptr){
        cout<<q->val<<" ";
        q = q->next;
    }cout<<endl;
}


class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        //ListNode* ans = (ListNode*) malloc(sizeof(ListNode));
        ListNode* ans = new ListNode;
        ListNode* r = ans;
        bool isFisrtNode = true;
        int carry = 0;//进位标识符
        while(l1 != nullptr && l2 != nullptr){
            int tmpValue = carry;
            carry = 0;
            tmpValue += l1->val + l2->val;
            if(tmpValue > 9){
                carry = tmpValue/10;
                tmpValue %= 10;
            }else{}
            if(!isFisrtNode){
                ListNode* s = new ListNode;
                s->val = tmpValue;
                r->next = s;
                r = s;
            }else{
                r->val = tmpValue;
                isFisrtNode = false;
            }
            l1 = l1->next;
            l2 = l2->next;
        }

//        printListNode(ans);

        while(l1 != nullptr){
            ListNode* s =  new ListNode;
            int tmpValue = l1->val;
            if(carry != 0){
                tmpValue += carry;
                carry = 0;
                if(tmpValue > 9){
                    carry = tmpValue/10;
                    tmpValue %= 10;
                }else{}
            }
            s->val = tmpValue;
            r->next = s;
            r = s;
            l1 = l1->next;
        }

        while(l2 != nullptr){
            ListNode* s = new ListNode;

            int tmpValue =  l2->val;
            if(carry != 0){
                tmpValue += carry;
                carry = 0;
                if(tmpValue > 9){
                    carry = tmpValue/10;
                    tmpValue %= 10;
                }else{}
            }

            s->val = tmpValue;
            r->next = s;
            r = s;
            l2 = l2->next;
        }

        if(carry != 0){
            ListNode* s = new ListNode;
            s->val = carry;
            r->next = s;
            r = s;
        }

        r->next = nullptr;

        return ans;
    }
};




int main() {
    //[2,4,9] [5,6,4,9] [7,0,4,0,1]
    //[2,4,3], l2 = [5,6,4]
    //输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
    //输出：[8,9,9,9,0,0,0,1]
    deque<int> node1{2,4,9};
    deque<int> node2{5,6,4,9};
    ListNode* p1 = createListNode(node1);
//    printListNode(p1);
    ListNode* p2 = createListNode(node2);
//    printListNode(p2);


    Solution s;
    ListNode* ans = s.addTwoNumbers(p1, p2);
    printListNode(ans);
    }