LintCode 167: Add Two Numbers (经典链表题)

最新推荐文章于 2021-07-18 12:55:44 发布

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本文介绍了一种算法，用于解决两个由链表表示的数字相加的问题。每个链表节点包含一个数字，数字以逆序方式存储。文章详细解释了如何创建一个函数来实现两个链表的数字相加，并返回结果作为新的链表。

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Add Two Numbers
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.

Return 2->1->9. That is 912.

Given 3->1->5 and 5->9->2, return 8->0->8.```

代码如下:

/**

Definition of singly-linked-list:
class ListNode {
public:
```
int val;
```
```
ListNode *next;
```
```
ListNode(int val) {
```
```
   this->val = val;
```
```
   this->next = NULL;
```
```
}
```
}
*/

class Solution {
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode * addLists(ListNode * l1, ListNode * l2) {
ListNode * node = new ListNode(0);
ListNode * dummyHead = node;
int carry = 0;
while(l1 && l2) {
ListNode *temp;
int sum = l1->val + l2->val + carry;
if (sum >= 10) {
carry = 1;
temp = new ListNode(sum % 10);
} else {
carry = 0;
temp = new ListNode(sum);
}
node->next = temp;
node = node->next;
l1 = l1->next;
l2 = l2->next;
}

    ListNode *l;
    if (l1) l = l1;
    else if (l2) l = l2;
    else {
        if (carry) node->next = new ListNode(1);
        return dummyHead->next;
    }

    while (l) {
        ListNode *temp;
        int sum = l->val + carry;
        if (sum >= 10) {
            carry = 1;
            temp = new ListNode (sum % 10);
        } else {
            carry = 0;
            temp = new ListNode (sum);
        }
        node->next = temp;
        node = node->next;
        l = l->next;
    }
    
    if (carry) node->next = new ListNode(1);
    
    return dummyHead->next;
}

};