LeetCode2-Add Two Numbers(C++)

Description

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order 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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

AC代码

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* t1 = l1;
        ListNode* t2 = l2;
        ListNode* l3 = new ListNode(0);   //l3用来保存结果
        ListNode* t3 = l3;   // 这里主要是为了记住l3的头指针的位置，避免的l3->next使得头指针找不回来了

        int val1 = 0;
        int val2 = 0;   // 用于表示当前正在计算的数字
        int sum = 0;   // 用户表示中间的计算结果
        int carry = 0;  // 用来表示进位，即sum/10, 如果carry=0表示没有进位，=1代表有进位。
        int dt = 0;   // 用于表示刨去进位后留下的数字，即sum%10;
        while(t1 != nullptr || t2 != nullptr){
            val1 = t1->val;
            val2 = t2->val;

            sum = val1 + val2 + carry;

            carry = sum / 10;
            dt = sum % 10;

            ListNode* newNode = new ListNode(dt);
            t3->next = newNode;
            t3 = t3->next;

            if(t1->next != nullptr && t2->next != nullptr) {
                t1 = t1->next;
                t2 = t2->next;
            } else if(t1->next == nullptr && t2->next != nullptr) {
                t3->next = t2->next;
                break;
            } else if(t2->next == nullptr && t1->next != nullptr) {
                t3->next = t1->next;
                break;
            } else {
                break;
            }

        }
        /*
            以下这部分是为了处理两个数组不等长的情况。有两个测试用例：
            1， [9,9]+[1]=[0,0,1]
            2， [8,9,9]+[2]=[0,0,0,1]
        */
        while(t3->next != nullptr) {
            int temp = t3->next->val;
            t3->next->val = (temp + carry) % 10;
            carry = (temp + carry) / 10;
            t3 = t3->next;
        }
        if(carry>0) {
            t3->next = new ListNode(1);
        }
        return l3->next;
    }
};

测试代码

int main() {
    ListNode* l1 = new ListNode(2);
    ListNode* l2 = new ListNode(4);
    ListNode* l3 = new ListNode(3);
    ListNode* l4 = new ListNode(5);
    ListNode* l5 = new ListNode(6);
    ListNode* l6 = new ListNode(4);

    l1->next = l2;
    l2->next = l3;
    l4->next = l5;
    l5->next = l6;

    Solution s;
    ListNode* result1 = s.addTwoNumbers(l1, l4);
    while(result1 != nullptr) {
        std::cout << result1->val << " ";
        result1 = result1->next;
    }
    std::cout << std::endl;

    ListNode* l7 = new ListNode(1);
    ListNode* l8 = new ListNode(8);
    l7->next = l8;
    ListNode* l9 = new ListNode(0);
    ListNode* result2 = s.addTwoNumbers(l7, l9);
    while(result2 != nullptr) {
        std::cout << result2->val << " ";
        result2 = result2->next;
    }
    std::cout << std::endl;

    ListNode* l10 = new ListNode(9);
    ListNode* l11 = new ListNode(9);
    l10->next = l11;
    ListNode* l12 = new ListNode(1);
    ListNode* result3 = s.addTwoNumbers(l10, l12);
    while(result3 != nullptr) {
        std::cout << result3->val << " ";
        result3 = result3->next;
    }
    std::cout << std::endl;

    ListNode* l13 = new ListNode(8);
    ListNode* l14 = new ListNode(9);
    ListNode* l15 = new ListNode(9);
    l13->next = l14;
    l14->next = l15;
    ListNode* l16 = new ListNode(2);
    ListNode* result4 = s.addTwoNumbers(l13, l16);
    while(result4 != nullptr) {
        std::cout << result4->val << " ";
        result4 = result4->next;
    }
    std::cout << std::endl;
}

总结

本题的基本思路就是创建一个新的链表，然后逐位相加，设置两个变量carry和dt用于保存进位和当前位的结果。本题的难度是Middle，处理起来最棘手的莫过于要处理两个数组不等长的情况。