[LeetCode]2. Add Two Numbers用链表逆序存储的两个数相加

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

一个很自然的想法是先分别遍历两个链表，求得两个加数add1和add2，然后求得sum=add1+add2，最后再将sum用链表按位逆序存储。这种方法忽略了链表长度很大时数据并不能存储在int或其他整型类型里的问题。即此题应该注意是一个大数问题。

思路：建立一个新链表，然后把输入的两个链表从头往后遍历，每两个相加，添加一个新节点到新链表后面，注意需要处理进位问题。还有就是最高位的进位问题要最后特殊处理一下。时间复杂度O(n)。

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *res = new ListNode(-1);
        ListNode *cur = res;
        int carry = 0;
        while (l1 || l2) {
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;
            int sum = n1 + n2 + carry;
            carry = sum / 10;
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
            if (l1) l1 = l1->next;
            if (l2) l2 = l2->next;
        }
        if (carry) cur->next = new ListNode(1);
        return res->next;
    }
};

在CareerCup上的这道题还有个Follow Up，把链表存的数字方向变了，原来是表头存最低位，现在是表头存最高位。这样稍微麻烦一些。一种想法是既然存的数字方向相反，那么我们可以先分别将两个链表反转，即可得到和上述一样的问题，这个时间复杂度同样也是O(n)：

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        int carry = 0;
        l1 = reverseList(l1);
        l2 = reverseList(l2);
        while (l1 || l2) {
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;
            int sum = n1 + n2 + carry;
            carry = sum / 10;
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
            if (l1) l1 = l1->next;
            if (l2) l2 = l2->next;
        }
        if (carry) cur->next = new ListNode(1);
        return reverseList(dummy->next);
    }
    ListNode *reverseList(ListNode *head) {
        if (!head) return head;
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;
        ListNode *cur = head;
        while (cur->next) {
            ListNode *tmp = cur->next;
            cur->next = tmp->next;
            tmp->next = dummy->next;
            dummy->next = tmp;
        }
        return dummy->next;
    }
};

第二种想法是先分别计算出两个链表的长度，然后给短一点的链表前面补0，补到和另一个链表相同的长度。由于要从低位开始相加，而低位是链表的末尾，所以我们采用递归来处理，先遍历到链表的末尾，然后从后面相加，进位标示符carry用的是引用，这样保证了再递归回溯时值可以正确传递，每次计算的节点后面接上上一次回溯的节点，直到回到首节点完成递归。最后还是处理最高位的进位问题：

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int n1 = 0, n2 = 0, carry = 0;;
        n1 = getLength(l1);
        n2 = getLength(l2);
        if (n1 > n2) l2 = padList(l2, n1 - n2);
        if (n2 > n1) l1 = padList(l1, n2 - n1);
        ListNode *res = addTwoNumbersDFS(l1, l2, carry);
        if (carry == 1) {
            ListNode *tmp = new ListNode(1);
            tmp->next = res;
            res = tmp;
        }
        return res;
    }
    ListNode *addTwoNumbersDFS(ListNode *l1, ListNode *l2, int &carry) {
        if (!l1 && !l2) return NULL;
        ListNode *list = addTwoNumbersDFS(l1->next, l2->next, carry);
        int sum = l1->val + l2->val + carry;
        ListNode *res = new ListNode(sum % 10);
        res->next = list;
        carry = sum / 10;
        return res;
    }
    ListNode *padList(ListNode *list, int len) {
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        for (int i = 0; i < len; ++i) {
            cur->next = new ListNode(0);
            cur = cur->next;
        }
        cur->next = list;
        return dummy->next;
    }
    int getLength(ListNode *list) {
        ListNode *cur = list;
        int res = 0;
        while (cur) {
            ++res;
            cur = cur->next;
        }
        return res;
    }
};