两数相加

2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

末尾置零法

关键点在于如何处理所给的两个链表长度不一致的情况。

末尾置零法的关键在于将短的链表长度用0补充直到长度和长度链表的长度相等。这样写代码比较简洁。

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    //创建头结点
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        //补充链表
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

链表法

链表法相比于末尾置零法，稍微复杂一点，需要判断l1，l2的长度。

若l1比l2长的话，只需根据l1和进位的情况，判断l1是否要进行修改。如果不需要进行修改，则只需将结果链表的下一个结点指向l1的剩余子链表的第一个结点即可。l2同理。

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

    int carry = 0;
   //创建头结点
    int temp = l1.val + l2.val;
    if(temp >= 10){
        temp = temp % 10;
        carry ++;
    }
    ListNode l3 = new ListNode(temp);
    ListNode l4 = l3; // l4 作为l3的头指针
    while(null != l1.next && null != l2.next){
        l1 = l1.next;
        l2 = l2.next;
        if(carry == 1){
            temp = l1.val + l2.val + carry;
            carry --;
        } else{
            temp = l1.val + l2.val;
        }
        if(temp >= 10){
            temp = temp % 10;
            carry ++;
        }
        l3.next = new ListNode(temp);
        l3 = l3.next;
    }

    // 如果l1长了，就把l1附在l3的后面
    while(null != l1.next){
        l1 = l1.next;
        if(carry == 1 ){
            temp = l1.val + carry;
            if(temp >= 10){
                temp = temp % 10;
                carry ++;
            }
            l3.next = new ListNode(temp);
            carry --;
        }else{
            l3.next = new ListNode(l1.val);
            l3.next.next = l1.next;
            break;
        }
        l3 = l3.next;
    }
    // 如果l2长了，就把l2附在l3的后面
    while(null != l2.next){
        l2 = l2.next;
        if(carry == 1 ){
            temp = l2.val + carry;
            if(temp >= 10){
                temp = temp % 10;
                carry ++;
            }
            l3.next = new ListNode(temp);
            carry --;
        }else{
            l3.next = new ListNode(l2.val);
            l3.next.next = l2.next;
        }
        l3 = l3.next;
    }
    //如果两者长度相等，有进位的话就附加一个进位
    if(carry == 1){
        l3.next = new ListNode(carry);
    }
    return l4;
}