【LeetCode】2. Add Two Numbers 解题报告

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转载请注明出处：http://blog.csdn.net/crazy1235/article/details/52914703

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Subject

出处：https://leetcode.com/problems/add-two-numbers/

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

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Explain

有两个链表，它们表示逆序的两个非负数。计算出两个数的和之后，同样逆序输出作为一个链表。

需要注意一点：有进位

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Solution

solution 1

便利两个链表，以此相加，把进位的数字计入下一组相加的数字中。

需要注意一种情况，比如{5} ， {5}两个列表，输出的结果应该是{0， 1}。

/**
     * 57ms 

     * 将个数较少的列表后面补上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;
    }

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solution 2

递归方式。

/**
     * 递归方式 -- 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;
    }

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solution 3

摘自『九章算法』的答案。

将过程分为三种情况。现将两个链表等长度对应的数字之和计算出来，然后两个链表有一个为空了。此时分两种情况（类似情况）。如果l1不为空，则将l1链表的值与“进位”相加再取余做为下一个结果元素的值。

/**
     * 九章算法答案 

     * 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;
    }

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bingo~~