兩個鏈表表示的數字相加 Add Two Numbers II

問題：

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

解決：

① 反轉兩個鏈表，相加以後再反轉回來。

class Solution { //54ms
    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode p1 = reverse(l1);
        ListNode p2 = reverse(l2);
        int carry = 0;
        ListNode header = new ListNode(-1);
        ListNode cur = header;
        while(p1 != null || p2 != null || carry != 0){
            int x = p1 == null ? 0 : p1.val;
            int y = p2 == null ? 0 : p2.val;
            int sum = x + y + carry;
            ListNode tmp = new ListNode(sum % 10);
            carry = sum / 10;
            tmp.next = cur.next;
            cur.next = tmp;
            cur = header;
            if(p1 != null) p1 = p1.next;
            if(p2 != null) p2 = p2.next;
        }
        return header.next;
    }
    public static ListNode reverse(ListNode head){
        if (head == null || head.next == null) return head;
        ListNode header = new ListNode(-1);
        ListNode cur = head;
        while(cur != null){
            ListNode next = cur.next;
            cur.next = header.next;
            header.next = cur;
            cur = next;
        }
        return header.next;
    }
}

② 不反轉，原地相加。首先計算長度，而後直接將對應的各位相加，最終，再掃描一次鏈表，更新每一個節點的數值，加上進位。

class Solution {//54ms
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null || l2 == null) return null;
        int len1 = 0;
        int len2 = 0;
        ListNode  p1 = l1;
        ListNode p2 = l2;
        while(p1 != null){//計算鏈表的長度
            p1 = p1.next;
            len1 ++;
        }
        while(p2 != null){
            p2 = p2.next;
            len2 ++;
        }
        int len = Math.max(len1,len2);
        ListNode pre = null;
        ListNode head;
        head = len1 > len2 ? l1 : l2;
        while(len != 0){
            int x = 0;
            int y = 0;
            if(len <= len1){
                x = l1.val;
                l1 = l1.next;
            }
            if(len <= len2){
                y = l2.val;
                l2 = l2.next;
            }
            len --;
            head.val = x + y;
            head.next = pre;
            pre = head;
            head = len1 > len2 ? l1 : l2;
        }
        head = pre;
        pre = null;
        int carry = 0;
        while(head != null){
            int cur = head.val + carry;
            head.val = cur % 10;
            carry = cur / 10;
            ListNode temp = head.next;
            head.next = pre;
            pre = head;
            head = temp;
        }
        if(carry != 0){
            ListNode result = new ListNode(carry);
            result.next = pre;
            return result;
        }
        return pre;
    }
}