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
刚开始ac的代码如下:简直是啰嗦到家
就是两个链表从头到尾非空的情况下往前计算
提防着进位等等 还要处理最后一个为空的边界条件
运行效率简直惨不忍睹:先贴这儿。
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { int carry = 0;//进位 ListNode* result = l1; while (l1->next != NULL && l2->next != NULL) { l1->val = l1->val + l2->val + carry; if (l1->val >= 10) { l1->val -= 10; carry = 1; } else carry = 0; l1 = l1->next; l2 = l2->next; } l1->val = l1->val + l2->val + carry; if (l1->val >= 10) { l1->val -= 10; carry = 1; } else carry = 0; if (l1->next == NULL && l2->next == NULL) { if (carry == 1) { ListNode* newNode = new ListNode(1); l1->next = newNode; } } else { if (l1->next == NULL && l2->next != NULL)//l2非空 { l1->next = l2->next; l1 = l1->next; while (carry && l1->next!=NULL) { l1->val += carry; if (l1->val >= 10) { l1->val -= 10; carry = 1; } else carry = 0; l1 = l1->next; } if (carry ==1)//最后一位 { l1->val += carry; if (l1->val >= 10) { l1->val -= 10; ListNode* newNode = new ListNode(1);//新建结点 l1->next = newNode; carry = 1; } else carry = 0; } } else if (l1->next != NULL && l2->next == NULL) { l1 = l1->next; while (carry && l1->next != NULL) { l1->val += carry; if (l1->val >= 10) { l1->val -= 10; carry = 1; } else carry = 0; l1 = l1->next; } if (carry == 1)//最后一位 { l1->val += carry; if (l1->val >= 10) { l1->val -= 10; ListNode* newNode = new ListNode(1);//新建结点 l1->next = newNode; carry = 1; } else carry = 0; } } } return result; }
简洁到家啊~~~ 如下:
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode c1 = l1; ListNode c2 = l2; ListNode sentinel = new ListNode(0); ListNode d = sentinel; int sum = 0; while (c1 != null || c2 != null) { sum /= 10; if (c1 != null) { sum += c1.val; c1 = c1.next; } if (c2 != null) { sum += c2.val; c2 = c2.next; } d.next = new ListNode(sum % 10); d = d.next; } if (sum / 10 == 1) d.next = new ListNode(1); return sentinel.next; }
不光简洁 效率也还过得去