Subject

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

Explain

有兩個連結串列，它們表示逆序的兩個非負數。計算出兩個數的和之後，同樣逆序輸出作為一個連結串列。

需要注意一點：有進位

Solution

solution 1

便利兩個連結串列，以此相加，把進位的數字計入下一組相加的數字中。

需要注意一種情況，比如{5} ， {5}兩個列表，輸出的結果應該是{0， 1}。

/**
     * 57ms <br />
     * 將個數較少的列表後面補上0
     * 
     * @param l1
     * @param l2
     * @return
 

     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null || l2 == null) {
            return null;
        }
        ListNode temp = new ListNode(0);
        ListNode result = temp;

        int value1 = 0;
        int value2 = 0;
        while (l1 != null && l2 != null
 
) {
            value2 = (l1.val + l2.val + value1) % 10;
            value1 = (l1.val + l2.val + value1) / 10;

            temp.next = new ListNode(value2);
            l1 = l1.next;
            l2 = l2.next;
            temp = temp.next;
            if (l1 == null && l2 == null) {
                break;
            }
            if (l1 == null) {
                l1 = new ListNode(0);
            }
            if (l2 == null) {
                l2 = new ListNode(0);
            }
        }
        if (value1 != 0) {
            temp.next = new ListNode(value1);
        }
        return result.next;
    }

solution 2

遞迴方式。

/**
     * 遞迴方式 -- 56ms
     * 
     * @param l1
     * @param l2
     * @return
     */
    public ListNode addTwoNumbers2(ListNode l1, ListNode l2) {
        if (l1 == null || l2 == null) {
            return l1 == null ? l2 : l1;
        }
        int value = l1.val + l2.val;
        ListNode result = new ListNode(value % 10);
        result.next = addTwoNumbers2(l1.next, l2.next);
        if (value >= 10) {
            result.next = addTwoNumbers2(new ListNode(value / 10), result.next);
        }
        return result;
    }

solution 3

摘自『九章演算法』的答案。

將過程分為三種情況。現將兩個連結串列等長度對應的數字之和計算出來，然後兩個連結串列有一個為空了。此時分兩種情況（類似情況）。如果l1不為空，則將l1連結串列的值與“進位”相加再取餘做為下一個結果元素的值。

/**
     * 九章演算法答案 <br />
     * http://www.jiuzhang.com/solutions/add-two-numbers/
     * 
     * @param l1
     * @param l2
     * @return
     */
    public ListNode addTwoNumbers3(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;
    }

bingo~~