LeetCode OJ : 2 Add Two Numbers

Add Two Numbers

Total Accepted: 103820 Total Submissions: 487934 Difficulty: Medium

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

/**
 * 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) {
        ListNode *ListSum = NULL, *pNode = NULL, *tail = NULL;
        ListNode *pl1 = l1, *pl2 = l2;
        int sum, cnt = 0;
        
        while(pl1 != NULL || pl2 != NULL){
            if(pl1 == NULL) {                   //此時是因為連結串列pl2比較長
                sum = cnt + pl2->val;
            }
            else if(pl2 == NULL){               //此時是因為連結串列pl1比較長
                sum = cnt + pl1->val;
            }
            else if(pl1 != NULL && pl2 != NULL){
                sum = cnt + pl1->val + pl2->val;
            }
            
            cnt = 0;
            if(sum >= 10){
                cnt = 1;
                sum -= 10;
            }
            
            pNode = new struct ListNode(sum);
            if(ListSum == NULL){
                ListSum = pNode;
                tail = pNode;
            }
            else{
                tail->next = pNode;
                tail = pNode;
            }
            
            if(pl1 != NULL)pl1 = pl1->next;
            if(pl2 != NULL)pl2 = pl2->next;
        }
        
        if(cnt == 1){                           //考慮到最高位的兩位進位
             sum = cnt;
             pNode = new struct ListNode(sum);
             if(ListSum == NULL){
                ListSum = pNode;
                tail = pNode;
             }
             else{
                tail->next = pNode;
                tail = pNode;
             }
        }
        
        return ListSum;
    }
};

如果將結果存放在其中一個比較長的連結串列上面，效率應該會更高一些；

寫了一半，懶得寫了，大概思路就是：首先獲得兩個連結串列的長度，通過比較兩個連結串列的長度，來分情況；

如果最高位還是有進位的話，就動態分配一個結點來儲存進位1；

改天有時間再試試；