借助链表实现两个数的相加

链式列表的数字高位在前与低位在前

这篇博客主要讲了借助链表实现两个数字的相加。链表分别是高位数字在前和低位数字在前

• 高位在前
• 低位在前

---

高位在前

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

对于这个题目，我实现的时候记住了栈，首先便利这两个链表，将数字放入栈里，由于高位在前，所以使用栈的话，低位就在栈顶了，然后遍历栈，结果放入一个栈，这样就又将高位放到了栈顶，这时候将数字从栈里弹出建立链表，就实现了；这种做法虽然时间复杂度是O(n)，但是空间复杂度也是O(n)；比较好的方法应该是将链表反向，然后直接相加，但是我这样做的时候出现了一些状况，还没有将程序弄对，等以后做出来了再贴上代码吧。
下面贴上这道题的代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        struct ListNode* ne = new struct ListNode(0);
        int carry = 0;
        int sum = 0;
        struct ListNode* temp = ne;
        struct ListNode* ll1 = l1;
        struct ListNode* ll2 = l2;
        stack<int> first;
        stack<int> second;
        stack<int> third;
        while (ll1 != NULL) {
            first.push(ll1->val);
            ll1 = ll1->next;
        }
        while (ll2 != NULL) {
            second.push(ll2->val);
            ll2 = ll2->next;
        }

        while (first.size() > 0 || second.size() > 0) {
            sum = 0;
            if (first.size() > 0) {
                sum += first.top();
                first.pop();
            }
            if (second.size() > 0) {
                sum += second.top();
                second.pop();
            }
            sum += carry;
            carry = sum / 10;
            sum = sum % 10;
            third.push(sum);
        }

        if (carry == 1)
            third.push(1);
        while (third.size() > 1) {
            temp -> val = third.top();
            third.pop();
            temp -> next = new struct ListNode(0);
            temp = temp -> next;
        }
        temp -> val = third.top();
        third.pop();

        return ne;
    }
};

低位在前

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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

对于这个题目的话，比较简单，只需要正向便利这两个链表，然后把对应的数字相加，每次相加之后记录下这两个数字相加产生的进位用于下一次加法就好了，时间复杂度是0(n),没有什么技术含量，只需要在相加的时候注意一下对指针的操作就好了。
下面贴上代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        struct ListNode* ne = (struct ListNode*)malloc(sizeof(struct ListNode));
    int carry = 0;
    int sum = 0;
    struct ListNode* temp = ne;
    while (l1 != NULL || l2 != NULL) {
        sum = 0;
        if (l1 != NULL) {
            sum = sum + l1->val;
            l1 = l1->next;
        }
        if (l2 != NULL) {
            sum = sum + l2->val;
            l2 = l2->next;
        }
        sum = sum + carry;
        carry = sum / 10;
        sum = sum % 10;
        temp->val = sum;

        if (l1 != NULL || l2 != NULL) {
            temp->next = (struct ListNode*)malloc(sizeof(struct ListNode));
            temp = temp->next;
        }
    }
    if (carry == 1) {
        temp->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        temp = temp->next;
        temp->val = 1;
    }
    temp -> next = NULL;
    return ne;
    }
};