LeetCode 2. Add Two Numbers ( C ++ )
Posted On 2021 年 4 月 21 日
題目為給定兩個鏈結串列,進行加法,回傳他們計算後的值(同樣以鏈結串列回傳),數字由左而右依序進位。
原文題目如下
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]
我採用的策略為,在較長的鏈結串列進行計算,結束後直接回傳該串列,以減少記憶體的使用量。
第一步先計算兩個鏈結串列的長度,並把 mainptr 指到較長的那個, optr 指到較短的那個,以 carry 變數計算進位值。
下方為我的程式碼
class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode *mainptr, *optr, *run; int carry = 0; if( CountList(l1) > CountList(l2) ){ mainptr = l1; optr = l2; } else{ mainptr = l2; optr = l1; } run = mainptr; while(optr != NULL || carry!=0){ if(optr != NULL){ carry += optr->val; optr = optr->next; } if(run != NULL){ carry += run->val; run->val = carry % 10; if(run->next != NULL){ run = run->next; } else{ if(carry / 10){ run->next = new ListNode(carry/10); run = run->next; return mainptr; } } } carry = carry / 10; } return mainptr; } private: int CountList(ListNode *input){ ListNode *temp = input; int counter = 0; while(temp!=NULL){ counter ++; temp = temp->next; } return counter; } };
下方為時間與空間之消耗
Runtime: 28 ms, faster than 62.33% of C++ online submissions for Add Two Numbers.
Memory Usage: 70.9 MB, less than 91.20% of C++ online submissions for Add Two Numbers.
下一題連結 : LeetCode 3. Longest Substring Without Repeating Characters ( C )-Medium