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代碼

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* t1 = l1;
        ListNode* t2 = l2;
        ListNode* l3 = new ListNode(0);   //l3用來保存結果
        ListNode* t3 = l3;   // 這裏主要是爲了記住l3的頭指針的位置，避免的l3->next使得頭指針找不回來了

        int val1 = 0;
        int val2 = 0;   // 用於表示當前正在計算的數字
        int sum = 0;   // 用戶表示中間的計算結果
        int carry = 0;  // 用來表示進位，即sum/10, 如果carry=0表示沒有進位，=1代表有進位。
        int dt = 0;   // 用於表示刨去進位後留下的數字，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;
            }

        }
        /*
            以下這部分是爲了處理兩個數組不等長的情況。有兩個測試用例：
            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;
    }
};

測試代碼

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;
}

總結

本題的基本思路就是創建一個新的鏈表，然後逐位相加，設置兩個變量carry和dt用於保存進位和當前位的結果。本題的難度是Middle，處理起來最棘手的莫過於要處理兩個數組不等長的情況。