LeedCode Add Two Numbers
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题目:
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
中文:
两个数字求和,数字用链表表示,每一个结点代表一位。链表顺序与数字顺序相反,即表头存放数字的最低位。
思路:
思路非常简单,利用两个指针分别遍历两个链表,并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可,若最后有进位需要添加一位。代码:
- /**
- * Definition for singly-linked list.
- * struct ListNode {
- * int val;
- * ListNode *next;
- * ListNode(int x) : val(x), next(NULL) {}
- * };
- */
- class Solution {
- public:
- ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
- int carry = 0; //是否进为
- ListNode* tail = new ListNode(0);
- ListNode* ptr = tail;
- while(l1 != NULL || l2 != NULL){ //当l1或l2不为空时
- int val1 = 0;
- if(l1 != NULL){
- val1 = l1->val;
- l1 = l1->next;
- }
- int val2 = 0;
- if(l2 != NULL){
- val2 = l2->val;
- l2 = l2->next;
- }
- int tmp = val1 + val2 + carry;
- ptr->next = new ListNode(tmp % 10); //ptr的下一个节点,新的节点,值为tem%10
- carry = tmp / 10; //进位
- ptr = ptr->next;
- }
- if(carry == 1){ //如果最后还有进位,必定是1
- ptr->next = new ListNode(1);
- }
- return tail->next;
- }
- };
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