445. Add Two Numbers II

问题描述：

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

解题思路：

由于是按照正常顺序排序，即最高位在最前面。

但是加法要从最低位开始。

所以我们可以用栈存储每一个节点。

将栈顶元素相加。

直至两栈为空。

空间复杂度O(m+n)

时间复杂度O(max(m+n))

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) {
        if(!l1) return l2;
        if(!l2) return l1;
        
        stack<ListNode*> s1, s2;
        ListNode* cur1 = l1;
        ListNode* cur2 = l2;
        
        while(cur1){
            s1.push(cur1);
            cur1 = cur1->next;
        }
        while(cur2){
            s2.push(cur2);
            cur2 = cur2->next;
        }
        
        ListNode* last = NULL;
        int carry = 0;
        while(!s1.empty() && !s2.empty()){
            int val = s1.top()->val + s2.top()->val + carry;
            s1.pop();
            s2.pop();
            carry = val / 10;
            val %= 10;
            ListNode* cur = new ListNode(val);
            cur->next = last;
            last = cur;
        }
        
        while(!s1.empty()){
            int val = s1.top()->val + carry;
            carry = val / 10;
            val %= 10;
            s1.pop();
            ListNode* cur = new ListNode(val);
            cur->next = last;
            last = cur;
        }
        
        while(!s2.empty()){
            int val = s2.top()->val + carry;
            carry = val / 10;
            val %= 10;
            s2.pop();
            ListNode* cur = new ListNode(val);
            cur->next = last;
            last = cur;
        }
        
        if(carry != 0){
            ListNode* cur = new ListNode(carry);
            cur->next = last;
            last = cur;
        }
        
        return last;        
        
    }
};

换一种写法：

/**
 * 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) {
        if(!l1) return l2;
        if(!l2) return l1;
        
        stack<ListNode*> s1, s2;
        ListNode* cur1 = l1;
        ListNode* cur2 = l2;
        
        while(cur1){
            s1.push(cur1);
            cur1 = cur1->next;
        }
        while(cur2){
            s2.push(cur2);
            cur2 = cur2->next;
        }
        
        ListNode* last = NULL;
        int carry = 0;
        while(!s1.empty() || !s2.empty() || carry != 0){
            int val1 = 0, val2 = 0;
            if(!s1.empty()){
                val1 = s1.top()->val;
                s1.pop();
            }
            if(!s2.empty()){
                val2 = s2.top()->val;
                s2.pop();
            }
            int val = val1 + val2 + carry;
            carry = val / 10;
            val %= 10;
            ListNode* cur = new ListNode(val);
            cur->next = last;
            last = cur;
        }
        
        return last;        
        
    }
};