167. Add Two Numbers【easy】

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

 

Example

Given 7->1->6 + 5->9->2. That is, 617 + 295.

Return 2->1->9. That is 912.

Given 3->1->5 and 5->9->2, return 8->0->8.

 

题意

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

 

解法一：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addLists(ListNode l1, ListNode l2) {
        if(l1 == null && l2 == null) {
            return null;
        }
            
        ListNode head = new ListNode(0);
        ListNode point = head;
        int carry = 0;
        while(l1 != null && l2!=null){
            int sum = carry + l1.val + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum / 10;
            l1 = l1.next;
            l2 = l2.next;
            point = point.next;
        }
        
        while(l1 != null) {
            int sum =  carry + l1.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l1 = l1.next;
            point = point.next;
        }
        
        while(l2 != null) {
            int sum =  carry + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l2 = l2.next;
            point = point.next;
        }
        
        if (carry != 0) {
            point.next = new ListNode(carry);
        }
        return head.next;
    }
}

中规中矩的解法

 

解法二：

public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;

        int carry = 0;
        for (ListNode i = l1, j = l2; i != null || j != null; ) {
            int sum = carry;
            sum += (i != null) ? i.val : 0;
            sum += (j != null) ? j.val : 0;

            tail.next = new ListNode(sum % 10);
            tail = tail.next;

            carry = sum / 10;
            i = (i == null) ? i : i.next;
            j = (j == null) ? j : j.next;
        }

        if (carry != 0) {
            tail.next = new ListNode(carry);
        }
        return dummy.next;
    }
}

比较简明的写法，且使用了dummy节点，参考@NineChapter 的代码

 

解法三：

public class Solution {
    /*
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    public ListNode addLists(ListNode l1, ListNode l2) {      
        ListNode root = new ListNode(-1);
        ListNode result = root;
        int carry = 0;        
        
        while( l1 != null || l2 != null || carry == 1){
            int value = 0;
            if(l1 != null){
                value += l1.val;
                l1 = l1.next;
            }
            if( l2 != null){
                value += l2.val;
                l2 = l2.next;
            }
            
            value += carry;            
            root.next = new ListNode(value % 10);
            carry = value / 10;
            
            root = root.next;
      
        }
        
        return result.next;
    }
}

 

解法四：

class Solution {
public:
    ListNode* addLists(ListNode* l1, ListNode* l2) {
        ListNode *head = new ListNode(0);
        ListNode *ptr = head;
        int carry = 0;
        while (true) {
            if (l1 != NULL) {
                carry += l1->val;
                l1 = l1->next;
            }
            if (l2 != NULL) {
                carry += l2->val;
                l2 = l2->next;
            }
            ptr->val = carry % 10;
            carry /= 10;
            // 当两个表非空或者仍有进位时需要继续运算，否则退出循环
            if (l1 != NULL || l2 != NULL || carry != 0) {
                ptr = (ptr->next = new ListNode(0));
            } else {
                break;
            }
        }
        return head;
    }
};

非常简明的代码，参考@NineChapter 的代码