【LeetCode】2. Add Two Numbers 解題報告

撰寫於2022-06-20閱讀次數

2. Add Two Numbers / Medium

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 contains a single digit. Add the two numbers and return the sum as a linked list.

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

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

  • The number of nodes in each linked list is in the range [1, 100].
  • 0 <= Node.val <= 9
  • It is guaranteed that the list represents a number that does not have leading zeros.

Solution

思路

想法其實很簡單，就跟加法器一樣，兩個值相加，算出是否需要進位，放回到其中一條 list。
如果其中一條沒了而另一條還有，就繼續加。最後要記得判斷是否還有 carry，有的話就要創一個新的節點。

效能

Complexity

  • Time Complexity: O(M+N), where M and N is the length of list1 and list2 respectively.
  • Space Complexity: O(1)

LeetCode Result

Code

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/**
 * Definition for singly-linked list.
 * 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) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *dummy = new ListNode(0);
        dummy->next = l1;
        ListNode *p = l1, *q = l2, *pre;
        int carry = 0;
        while(p && q) {
            int x = p ? p->val : 0;
            int y = q ? q->val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            p->val = sum % 10;
            pre = p;
            p = p->next;
            q = q->next;
        }
        
        if(q) {
            pre->next = q;
            p = pre->next;
        }
        
        while(p) {
            p->val = p->val + carry;
            carry = p->val / 10;
            p->val = p->val % 10;
            pre = p;
            p = p->next;
        }
        
        if(carry > 0) pre->next = new ListNode(1);
        return dummy->next;
    }
};