算法设计Week1 LeetCode Algorithms Problem #2 Add Two Numbers
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题目描述:
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
使用C++解题给出的ListNode
结构体定义:
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} };
这道题本身并不难,将数字反序存放到链表里也减弱了本题的难度。最一般的情况就是两个链表等长,每一位对应相加直到链表为空,注意考虑进位。此外,还要考虑两个链表不等长(一个链表为空可合并考虑)的情况,这里需要特殊考虑的是连续进位的情况:如[5,9,9,9]和[8,3]应得到[3,3,0,0,1]。在两个链表不等长时,必有一个链表先为空,这时可以直接将不为空的链表对应的后半部分接到结果上,然后再考虑进位,如若进位则一直进行计算,直到进位为0。
解法一(Time Limit Exceeded):
这里卡了很久,算法应该是没有问题的,代码应该也没有问题,但就是超时。最后改了很久,超时的原因应该是在第一个循环里对result是否为空的判断引起的计算缓慢(虽然还是想不明白为什么),最后将这部分去掉并做一些修改就通过了。
class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* result; ListNode* result_head; int carry = 0; int temp; // digits that l1 and l2 both contain (low digits) while(l1 && l2){ // 这一部分就是超时的“罪魁祸首” if(result == NULL){ result = new ListNode(0); result_head = result; } else{ result->next = new ListNode(0); result = result->next; } // 这一部分就是超时的“罪魁祸首” temp = l1->val + l2->val + carry; result->val = temp % 10; carry = temp / 10; l1 = l1->next; l2 = l2->next; } // give the longer number remains to result result->next = l1 ? l1 : l2; // deal with carry while(carry){ if(result->next == NULL){ result->next = new ListNode(1); carry = 0; }else{ result = result->next; temp = result->val + carry; carry = temp / 10; result->val = temp % 10; } } return result_head; }};
解法二:
为了改正上面的错误,将标注部分去掉后需要对程序进行微调。采取的方法是在result
前加一个节点,最后返回result_head->next
,这样就可以避免在循环的开头不断对result
是否为空进行判断了。最后用时52ms。
class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* result = new ListNode(0); // 修改处 ListNode* result_head = result; // 修改处 int carry = 0; int temp; // digits that l1 and l2 both contain (low digits) while(l1 && l2){ result->next = new ListNode(0); result = result->next; temp = l1->val + l2->val + carry; result->val = temp % 10; carry = temp / 10; l1 = l1->next; l2 = l2->next; } // give the longer number remains to result result->next = l1 ? l1 : l2; // deal with carry while(carry){ if(result->next == NULL){ result->next = new ListNode(1); carry = 0; }else{ result = result->next; temp = result->val + carry; carry = temp / 10; result->val = temp % 10; } } return result_head->next; }};
解法三
一个十分简洁的版本,仅12行。为了使得代码更加简洁,将所有情况一起考虑,最后仅用了一个循环就解决了问题,虽然代码行数更少,但由于增加了计算时所需的比较次数,整体代码运行较慢,用时62ms。
class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* result = new ListNode(0); ListNode* result_head = result; int carry = 0; int sum = 0; while(l1 || l2 || carry){ sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry; result->next = new ListNode(sum % 10); carry = sum / 10; result = result->next; l1 = l1 ? l1->next : l1; l2 = l2 ? l2->next : l2; } return result_head->next; }};
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