[leetcode]LeetCode刷题的日子--Add Two Numbers

刷题第二天。今天下班其实很早，但是一晚上都在学rust的Option，Some，Box的用法，用来刷题的时间基本都没有了，所以只刷了一道。

题目如下：

You are given two non-empty linked 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.

题目翻译：

一个不为空的单链表表示一个非负整数，其中头部位低位，尾部为高位，计算两个数相加的结果。

题目思路：

从头部开始遍历，逐级相加，记得考虑进位

代码如下：

struct ListNode {
    pub val: i32,
    pub next: Option<Box<ListNode>>
}

impl ListNode {
    #[inline]
    fn new(val: i32) -> Self {
        ListNode {
            next: None,
            val
        }
    }
}

fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    let mut ret = Some(Box::new(ListNode::new(0)));

    let (mut p1, mut p2, mut carry) = (l1, l2, 0);
    let mut pos = ret.as_mut();
    while p1.is_some() || p2.is_some() {
        let mut sum = carry;

        if let Some(v) = p1 {
            sum += v.val;
            p1 = v.next;
        }

        if let Some(v) = p2 {
            sum += v.val;
            p2 = v.next;
        }

        if sum >= 10 {
            carry = 1;
            sum = sum - 10;
        } else {
            carry = 0;
        }

        let v = pos.unwrap();
        v.next = Some(Box::new(ListNode::new(sum)));
        pos = v.next.as_mut();
    }

    if carry > 0 {
        let v = pos.unwrap();
        v.next = Some(Box::new(ListNode::new(1)));
    }

    return ret.unwrap().next;
}

#[test]
fn test_add_two_numbers() {
    assert_eq!(1, 1);

    let mut l1 = Box::new(ListNode::new(2));
    l1.next = Some(Box::new(ListNode::new(4)));
    l1.next.as_mut().unwrap().next= Some(Box::new(ListNode::new(7)));

    let mut l2 = Box::new(ListNode::new(5));
    l2.next = Some(Box::new(ListNode::new(6)));
    l2.next.as_mut().unwrap().next = Some(Box::new(ListNode::new(4)));

    let ret = add_two_numbers(Some(l1), Some(l2));

    let expected_ret = [7, 0, 2, 1];
    let mut pos = ret;
    for i in (0..expected_ret.len()) {
        assert_eq!(pos.is_some(), true);
        if !pos.is_some() {
            break;
        }

        if let Some(v) = pos {
            assert_eq!(v.val, expected_ret[i]);
            pos = v.next;
        }

    }
}