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2014
11-19

LeetCode-Add Two Numbers[链表]

Add Two Numbers

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

标签: Linked List Math
分析

跟Add Binary(见 \S \ref{sec:add-binary})很类似

代码1

// LeetCode, Add Two Numbers
// 跟Add Binary 很类似
// 时间复杂度O(m+n),空间复杂度O(1)
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode dummy(-1); // 头节点
        int carry = 0;
        ListNode *prev = &dummy;
        for (ListNode *pa = l1, *pb = l2;
             pa != nullptr || pb != nullptr;
             pa = pa == nullptr ? nullptr : pa->next,
             pb = pb == nullptr ? nullptr : pb->next,
             prev = prev->next) {
            const int ai = pa == nullptr ? 0 : pa->val;
            const int bi = pb == nullptr ? 0 : pb->val;
            const int value = (ai + bi + carry) % 10;
            carry = (ai + bi + carry) / 10;
            prev->next = new ListNode(value); // 尾插法
        }
        if (carry > 0)
            prev->next = new ListNode(carry);
        return dummy.next;
    }
};

  1. 有两个重复的话结果是正确的,但解法不够严谨,后面重复的覆盖掉前面的,由于题目数据限制也比较严,所以能提交通过。已更新算法

  2. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
    O(n) = O(n-1) + O(n-2) + ….
    O(n-1) = O(n-2) + O(n-3)+ …
    O(n) – O(n-1) = O(n-1)
    O(n) = 2O(n-1)

  3. 我没看懂题目
    2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5
    我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
    第二个是7 0 6 -1 1 -6 7输出是14 7 7
    不知道题目例子是怎么得出来的