LeetCode 445——兩數相加 II

1. 題目

2. 解答

2.1 方法一

在 LeetCode 206——反轉鏈表 和 LeetCode 2——兩數相加 的基礎上，先對兩個鏈表進行反轉，而後求出和後再進行反轉便可。

/**
 * 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) {
        
        // 先將兩個鏈表反序
        l1 = reverseList(l1);
        l2 = reverseList(l2);
        
        ListNode *head = new ListNode(0); // 創建哨兵結點
        ListNode *temp = head; 

        
        int carry = 0; // 保留進位
        int sum = 0;
            
        while(l1 || l2)
        {
            if (l1 && l2) // 兩個鏈表都非空
            {
                sum = l1->val + l2->val + carry;
                l1 = l1->next;
                l2 = l2->next;
            }
            else if (l1) // l2 鏈表爲空，只對進位和 l1 元素求和
            {
                sum = l1->val + carry;
                l1 = l1->next;
            }
            else // l1 鏈表爲空，只對進位和 l2 元素求和
            {
                sum = l2->val + carry;
                l2 = l2->next;
            } 
            
            // 求出和以及進位，將和添加到新鏈表中
            carry = sum >= 10 ? 1 : 0;
            sum = sum % 10;
            head->next = new ListNode(sum);
            head = head->next;
            
            if ( (l1 == NULL || l2 == NULL) && carry == 0 )
            {
                head->next = l1 ? l1 : l2;
                return reverseList(temp->next);
            }

        }
          
    
        if (carry) // 若最後一位還有進位，進位即爲最後一位的和
        {
            head->next = new ListNode(carry);    
        }
        head->next->next = NULL;
        

        return reverseList(temp->next);
        
    }
    
    ListNode* reverseList(ListNode* head) {
    
        if (head == NULL || head->next == NULL) return head; 
        else 
        { 
            ListNode * p1 = head; 
            ListNode * p2 = p1->next; 
            ListNode * p3 = p2->next; 
            
            while (p2) 
            { 
                p3 = p2->next; p2->next = p1; 
                p1 = p2; 
                p2 = p3; 
            } 
            head->next = NULL; 
            head = p1; 

            return head; 
        }

    }

};

2.2 方法二

先求出兩個鏈表的長度，而後對齊兩個鏈表，按照對應位分別求出每一位的和以及進位，最後從最低位也就是最右邊開始，將和與進位相加，新建節點在鏈表頭部插入便可。

例 1
l1	7	2	4	3
l2		5	6	4
和	7	7	0	7
進位	0	1	0	0
結果	7	8	0	7

例 2
l1		9	9	9
l2			9	9
和	0	9	8	8
進位	0	1	1	0
結果	1	0	9	8

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        
        int n1 = countListNodeNumber(l1);
        int n2 = countListNodeNumber(l2);
        
        ListNode* long_list = NULL;
        ListNode* short_list = NULL;
        int bigger_n = 0;
        int smaller_n = 0;
        
        if (n1 <= n2)
        {
            long_list = l2;
            short_list = l1;
            bigger_n = n2;
            smaller_n = n1;
        }
        else
        {
            long_list = l1;
            short_list = l2;
            bigger_n = n1;
            smaller_n = n2;   
        }
        int temp = bigger_n - smaller_n + 1;
        int sum_array[bigger_n + 1] = {0};
        int carry_array[bigger_n + 1] = {0};
        int sum = 0;
        int carry = 0;
        
        ListNode* long_temp = long_list;
        ListNode* short_temp = short_list;
        
        for (unsigned int i = 1; i <= bigger_n; i++)
        {
            carry = 0;
            if (i < temp)
            {
                sum = long_temp->val;
            }
            else
            {
                sum = long_temp->val + short_temp->val;
                if (sum >= 10)
                {
                    sum = sum % 10;
                    carry = 1;
                }
                short_temp = short_temp->next;
            }
            sum_array[i] = sum;
            carry_array[i-1] = carry;
            long_temp = long_temp->next;
        }
        
               
        ListNode* new_head = new ListNode(0);
        long_temp = new_head;
        
        for (unsigned int i = bigger_n; i > 0; i--)
        {
            // 在鏈表頭部進行插入
            sum = sum_array[i] + carry_array[i];
            if (sum >= 10)
            {
                sum = sum % 10;
                carry_array[i-1] = 1;
            }
            short_temp = new ListNode(sum);
            short_temp->next = long_temp->next;
            long_temp->next = short_temp;
        }
        
        sum = sum_array[0] + carry_array[0];
        if (sum)
        {
            short_temp = new ListNode(sum);
            short_temp->next = long_temp->next;
            long_temp->next = short_temp;
        }
        
        return new_head->next;

    }
    
    int countListNodeNumber(ListNode *head)
    {
        int node_num = 0;
        ListNode* slow = head; 
        ListNode* fast = head; 
        
        while(fast && fast->next) 
        {
            slow = slow->next; 
            fast = fast->next->next; 
            node_num++; 
        } 
        
        if (fast) 
        { 
            node_num = node_num * 2 + 1;
        } 
        else 
        { 
            node_num = node_num * 2; 
        }
        
        return node_num;
    }
    
    
};

2.3 方法三

將兩個鏈表的節點分別放入兩個棧中，若兩個棧都非空，拿兩個棧頂元素和進位，求出和以及新的進位；若其中一個棧爲空，則拿一個棧頂元素和進位，求出和以及新的進位。而後新建節點，在鏈表頭部進行插入，最後只用處理一下最高位的進位便可。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        
        stack<ListNode *> stack1;
        stack<ListNode *> stack2;
        
        while (l1)
        {
            stack1.push(l1);
            l1 = l1->next;
        }
        
        while (l2)
        {
            stack2.push(l2);
            l2 = l2->next;
        }
        
        int sum = 0;
        int carry = 0;
        
        ListNode *new_head = new ListNode(0);
        ListNode *temp = NULL;
        
        while (!stack1.empty() && !stack2.empty())
        {
            sum = stack1.top()->val + stack2.top()->val + carry;
            
            if (sum >= 10)
            {
                sum = sum % 10;
                carry = 1;
            }
            else
                carry = 0;
            
            temp = new ListNode(sum);
            temp->next = new_head->next;
            new_head->next = temp; 
            
            stack1.pop();
            stack2.pop();
        }
                

         while (!stack1.empty())
         {
            sum = stack1.top()->val + carry;

            if (sum >= 10)
            {
                sum = sum % 10;
                carry = 1;
            }
            else
                carry = 0;

            temp = new ListNode(sum);
            temp->next = new_head->next;
            new_head->next = temp; 

            stack1.pop();
         }
        

         while (!stack2.empty())
         {
            sum = stack2.top()->val + carry;

            if (sum >= 10)
            {
                sum = sum % 10;
                carry = 1;
            }
            else
                carry = 0;

            temp = new ListNode(sum);
            temp->next = new_head->next;
            new_head->next = temp; 

            stack2.pop();
         }
        
        if (carry)
        {
            temp = new ListNode(carry);
            temp->next = new_head->next;
            new_head->next = temp; 
        }
        
        return new_head->next;
    }   
};

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