Two Lists Sum

Source

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.

题解

一道看似简单的进位加法题，实则杀机重重，不信你不看答案自己先做做看。

首先由十进制加法可知应该注意进位的处理，但是这道题仅注意到这点就够了吗？还不够！因为两个链表长度有可能不等长！因此这道题的亮点在于边界和异常条件的处理，来瞅瞅我自认为相对优雅的实现。

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoLists(ListNode* l1, ListNode* l2) {
        ListNode dummy(0);
        ListNode *curr = &dummy;
        int carry = 0;

        while ((l1 != NULL) || (l2 != NULL) || (carry != 0)) {
            int l1_val = (l1 != NULL) ? l1->val : 0;
            int l2_val = (l2 != NULL) ? l2->val : 0;
            int sum = carry + l1_val + l2_val;
            carry = sum / 10;
            curr->next = new ListNode(sum % 10);

            curr = curr->next;
            if (l1 != NULL) l1 = l1->next;
            if (l2 != NULL) l2 = l2->next;
        }

        return dummy.next;
    }
};

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode curr = dummy;
        int carry = 0;

        while ((l1 != null) || (l2 != null) || (carry != 0)) {
            int l1_val = (l1 != null) ? l1.val : 0;
            int l2_val = (l2 != null) ? l2.val : 0;
            int sum = carry + l1_val + l2_val;
        // update carry
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);

            curr = curr.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }

        return dummy.next;
    }
}

源码分析

1. 迭代能正常进行的条件为(NULL != l1) || (NULL != l2) || (0 != carry), 缺一不可。
2. 对于空指针节点的处理可以用相对优雅的方式处理 - int l1_val = (NULL == l1) ? 0 : l1->val;

复杂度分析

没啥好分析的，时间和空间复杂度均为 O(max(L1,L2)).