【leetcode】2. Add Two Numbers(Python & C++)
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2. Add Two Numbers
题目链接
2.1 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
2.2 解题思路:
思路一:同时遍历两个链表,并把遍历的两个元素相加,创建新链表元素,记录和。如果其中一个链表遍历结束,则新链表直接连接另一个链表剩余元素。最后再处理新链表每个元素是否大于10,要进位的问题。
思路二:同思路一相同,但不同的是在遍历两个元素相加的时候,就判断是否要进位的问题,并用flag记录。这里要注意的是,当两个链表都遍历完,且最后一次计算flag为1,需要进位,则直接创建值为1的新节点连接到链表后面。
思路三:初始化sum值为0。不同于前两种思路遍历条件并不是两个链表非空,而是其中之一不空即可。进入到遍历时,将sum/10,获取进位值。然后依次判断链表是否为空,不空则其值依次与sum相加。新链表的节点值为sum%10。最后在遍历结束后,再判断依次sum/10,若为1,则创建值为1的节点连接到新链表后。
思路四:同思路三一样,不过写法上更加简洁。
2.3 C++代码:
1、思路一代码(72ms):
class Solution91 {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if (l1 == NULL && l2 == NULL) return NULL; if (l1 == NULL) return l2; if (l2 == NULL) return l1; ListNode *p1 = l1; ListNode *p2 = l2; ListNode *q = new ListNode(-1); ListNode *p = q; while (p1!=NULL && p2!=NULL) { ListNode *temp = new ListNode(p1->val + p2->val); p->next = temp; p = p->next; p1 = p1->next; p2 = p2->next; } if (p1!=NULL) p->next = p1; if (p2 != NULL) p->next = p2; int flag = 0; ListNode *t = q->next; while (t!=NULL) { t->val = t->val + flag; if (t->val>= 10) { t->val = t->val % 10; flag = 1; } else flag = 0; if (flag == 1 && t->next==NULL) { ListNode *m = new ListNode(1); t->next = m; break; } t = t->next; } return q->next; }};
2、思路二代码(46ms):
class Solution91_1 {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if (l1 == NULL && l2 == NULL) return NULL; if (l1 == NULL) return l2; if (l2 == NULL) return l1; ListNode *p1 = l1; ListNode *p2 = l2; ListNode *q = new ListNode(-1); ListNode *p = q; int flag = 0; while (p1 != NULL && p2 != NULL) { ListNode *temp = new ListNode(p1->val + p2->val + flag); if (temp->val >= 10) { temp->val = temp->val % 10; flag = 1; } else flag = 0; p->next = temp; p = p->next; if (flag == 1 && p1->next == NULL && p2->next==NULL) { ListNode *x = new ListNode(1); p->next = x; } p1 = p1->next; p2 = p2->next; } while (p1 != NULL) { p1->val = p1->val + flag; if (p1->val >= 10) { p1->val = p1->val % 10; flag = 1; } else flag = 0; p->next = p1; p = p->next; if (flag == 1 && p1->next==NULL) { ListNode *x = new ListNode(1); p->next = x; break; } p1 = p1->next; } while (p2 != NULL) { p2->val = p2->val + flag; if (p2->val >= 10) { p2->val = p2->val % 10; flag = 1; } else flag = 0; p->next = p2; p = p->next; if (flag == 1 && p2->next == NULL) { ListNode *x = new ListNode(1); p->next = x; break; } p2 = p2->next; } return q->next; }};
3、思路三代码(32ms):
class Solution91_2 {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if (l1 == NULL && l2 == NULL) return NULL; if (l1 == NULL) return l2; if (l2 == NULL) return l1; ListNode *p1 = l1; ListNode *p2 = l2; ListNode *q = new ListNode(-1); ListNode *p = q; int sum = 0; while (p1 != NULL || p2 != NULL) { sum = sum / 10; if (p1 != NULL) { sum = sum + p1->val; p1 = p1->next; } if (p2 != NULL) { sum = sum + p2->val; p2 = p2->next; } p->next = new ListNode(sum % 10); p = p->next; } if (sum / 10 == 1) p->next = new ListNode(1); return q->next; }};
4、思路四代码(29ms):
class Solution91_3 {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if (l1 == NULL && l2 == NULL) return NULL; if (l1 == NULL) return l2; if (l2 == NULL) return l1; ListNode *p1 = l1; ListNode *p2 = l2; ListNode *q = new ListNode(-1); ListNode *p = q; int flag = 0; while (p1 || p2 || flag) { int sum = (p1 ? p1->val : 0) + (p2 ? p2->val : 0) + flag; p->next = new ListNode(sum % 10); p = p->next; flag = sum / 10; p1 = p1 ? p1->next : p1; p2 = p2 ? p2->next : p2; } return q->next; }};
2.4 Python代码:
1、思路二代码(115ms):
class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ if l1==None and l2==None: return None if l1==None: return l2 if l2==None: return l1 q=ListNode(-1) p=q flag=0 p1=l1 p2=l2 while p1!=None and p2!=None: temp=ListNode(p1.val+p2.val+flag) if temp.val>=10: temp.val=temp.val%10 flag=1 else: flag=0 p.next=temp p=p.next if flag==1 and p1.next==None and p2.next==None: t=ListNode(1) p.next=t p1=p1.next p2=p2.next while p1!=None: p1.val=p1.val+flag if p1.val>=10: p1.val=p1.val%10 flag=1 else: flag=0 p.next=p1 p=p.next if flag==1 and p1.next==None: t=ListNode(1) p.next=t break p1=p1.next while p2!=None: p2.val=p2.val+flag if p2.val>=10: p2.val=p2.val%10 flag=1 else: flag=0 p.next=p2 p=p.next if flag==1 and p2.next==None: t=ListNode(1) p.next=t break p2=p2.next return q.next
2、思路三代码(132ms):
class Solution1(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ if l1==None and l2==None: return None if l1==None: return l2 if l2==None: return l1 q=ListNode(-1) p=q flag=0 p1=l1 p2=l2 while p1 or p2 or flag: sum=flag if p1: sum=sum+p1.val p1=p1.next if p2: sum=sum+p2.val p2=p2.next p.next=ListNode(sum%10) p=p.next flag=sum/10 return q.next
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