题目
You are given two linked non-empty 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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
题目要求的是将两条数字链相加,返回加完后的正确结果。由于是指针问题,所以在申请新内存方面要多考虑,有以下三个较为重要的情况需要考虑:
- 两条链长度不同时,不能忘了长的链后面的数也要加上。
- 当加到其中有一条链为空时,还要继续考虑进位的问题。例如:(1->2)+ (1->8->7->6)
- 当加到最后有新的进位时,需要再申请内存存数。例如:(1->2)+ (1->8->9->9)
由于题目说明为两条非空链,如果没有说明的话,我们还需要考虑当两条链同时为空或者有一条为空链的情况。
C++代码如下:
/**
* 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) {
ListNode* result, *head;
int carry = 0;
//首先确定和的头指针
int tmp = l1->val + l2->val;
result = new ListNode(tmp % 10);
head = result;
carry = tmp / 10;
l1 = l1->next;
l2 = l2->next;
//两条链不为空时一直相加
while (l1 != NULL && l2 != NULL) {
int tmp = l1->val + l2->val;
result->next = new ListNode((tmp + carry) % 10);
carry = (tmp + carry) / 10;
result = result->next;
l1 = l1->next;
l2 = l2->next;
}
/*
当加到有一条链为空时,结果需要加上另一条链剩下的数,
同时不能忘了在另一条链为空时,可能存在进位。
*/
if (l1 == NULL) {
while (l2 != NULL) {
result->next = new ListNode((l2->val + carry) % 10);
carry = (l2->val + carry) / 10;
result = result->next;
l2 = l2->next;
}
}
else {
while (l1 != NULL) {
result->next = new ListNode((l1->val + carry) % 10);
carry = (l1->val + carry) / 10;
result = result->next;
l1 = l1->next;
}
}
//考虑最后是否还有进位,有则申请行内存
if (carry) {
result->next = new ListNode(1);
}
return head;
}
};