Add Two Numbers

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本文介绍了一种通过链表实现两数相加的方法，包括算法解析、迭代和递归两种实现方式，以及如何处理进位情况。

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

Analysis:

1. 两个指针一起遍历，若有一个先为null，则在剩余的遍历中加0代替之。

2. 遍历结束之后，若还需进位，则应该再创建一个新的结点用于存储这个carry。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null) {
            return l2;
        }
        else if(l2 == null) {
            return l1;
        }

        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode res = new ListNode(-1);
        ListNode c3 = res;
        ListNode tail = null;
        int carry = 0;
        
        while(c1 != null || c2 !=null) {        // once null, add 0 instead
            int v1 = c1==null ? 0 : c1.val;
            int v2 = c2==null ? 0 : c2.val;
            c3.val = (v1+v2+carry) % 10;
            c3.next = new ListNode(-1);
            tail = c3;
            c3 = c3.next;
            carry = (v1+v2+carry) / 10;
            c1 = c1==null ? null : c1.next;
            c2 = c2==null ? null : c2.next;
        }
        
        if(carry == 1) {            // one more bit needs
            c3.val = carry;
        }
        else {
            tail.next = null;       // delete the last redundant bit
        }
        
        return res;
 
    }
}

Update 02/04/2014:

iterative:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null) return l2;
        if(l2 == null) return l1;

        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode res = new ListNode(-1);
        ListNode c3 = res;
        int carry = 0;
        
        while(c1 != null || c2 !=null) {        // once null, add 0 instead
            int v1 = c1==null ? 0 : c1.val;
            int v2 = c2==null ? 0 : c2.val;
            c3.next = new ListNode((v1+v2+carry) % 10);
            c1 = c1==null ? null : c1.next;
            c2 = c2==null ? null : c2.next;
            carry = (v1+v2+carry) / 10;
            c3 = c3.next;
        }
        
        if(carry==1) c3.next=new ListNode(carry);  // one more bit needs
        return res.next;
    }
}

recursive:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2, int carry) {
        if(l1==null && l2==null && carry==0) return null;
        int val = carry;
        if(l1!=null) val+=l1.val;
        if(l2!=null) val+=l2.val;
        ListNode result = new ListNode(val%10);
        result.next = addTwoNumbers(l1==null?null:l1.next, 
                                    l2==null?null:l2.next,
                                    val>=10?1:0);
        return result;
    }
    
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return addTwoNumbers(l1, l2, 0);
    }
}