Add Two Numbers

21% Accepted

You have two numbers represented by a linked list, where each node contains a single digit.
The digits are stored in reverse order, such that the 1's digit is at the head of the list.
Write a function that adds the two numbers and returns the sum as a linked list.

Have you met this question in a real interview? Yes
Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.

Return 2->1->9. That is 912.

Given 3->1->5 and 5->9->2, return 8->0->8.

Tags Expand

• Cracking The Coding Interview
• Linked List
• High Precision

解法一

• 使用了StringBuffer去解决问题，去避免了Linkedlist太长，导致int或者long溢出的问题
• Lintcode跑完所有case是1100ms左右
• Character.getNumericValue(res.charAt(i)) 将char转化为int
• 创建了string add 函数，以后可以自己调用

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        // write your code here
        if (l1 == null || l2 == null) {
            return null;
        }
        StringBuffer res = stringadd(getnum(l1), getnum(l2));
        int len = res.length();
        ListNode root = new ListNode(0);
        ListNode cur = root;
        for (int i = len - 1; i >= 0; i--) {
            cur.val = Character.getNumericValue(res.charAt(i));
            if (i != 0) {
                ListNode next = new ListNode(0);
                cur.next = next;
                cur = cur.next;
            }
        }
        return root;

    }
    public StringBuffer getnum(ListNode l1) {
        StringBuffer sum = new StringBuffer("");
        sum.append(l1.val);
        while (l1.next != null) {
            sum.insert(0,l1.next.val);
            l1 = l1.next;
        }
        return sum;
    }

    public StringBuffer stringadd(StringBuffer s1, StringBuffer s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        int i = 0;
        StringBuffer res = new StringBuffer("");
        int carry = 0;
        while (i < Math.max(len1, len2)) {

            int add1, add2;
            if (i < len1) {
                add1 = Character.getNumericValue(s1.charAt(len1 - 1 - i));
            } else {
                add1 = 0;
            }
            if (i < len2) {
                add2 = Character.getNumericValue(s2.charAt(len2 - 1 - i));
            } else {
                add2 = 0;
            }
            int sum = (add1 + add2 + carry) % 10;
            if ((add1 + add2 + carry) >= 10) {
                carry = 1;
            } else {
                carry = 0;
            }
            i++;
            res.insert(0, sum);
        }
        if (carry == 1) {
            res.insert(0, 1);
        }
        return res;
    }
}

解法二

• 直接计算，根本不必要去读取linkedlist数值，因为本来就翻转的，最前面的数就是个位数，其次是十位数，直接相加就可以了

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

        ListNode dummy = new ListNode(0);
        ListNode result = dummy;

        int carryBit = 0;
        while (l1 != null && l2 != null) {
            int newVal = l2.val + l1.val + carryBit;
            if (newVal >= 10) {
                newVal -= 10;
                carryBit = 1;
            } else {
                carryBit = 0;
            }
            result.next = new ListNode(newVal);
            result = result.next;
            l1 = l1.next;
            l2 = l2.next;
        }

        while (l1 != null) {
            int newVal = l1.val;
            if (carryBit != 0) {
                newVal = l1.val + carryBit;
            }
            if (newVal >= 10) {
                newVal -= 10;
                carryBit = 1;
            } else {
                carryBit = 0;
            }
            result.next = new ListNode(newVal);
            result = result.next;
            l1 = l1.next;
        }

        while (l2 != null) {
            int newVal = l2.val;
            if (carryBit != 0) {
                newVal = l2.val + carryBit;
            }
            if (newVal >= 10) {
                newVal -= 10;
                carryBit = 1;
            } else {
                carryBit = 0;
            }
            result.next = new ListNode(newVal);
            result = result.next;
            l2 = l2.next;
        }

        if (carryBit == 1) {
            result.next = new ListNode(carryBit);
        }

        return dummy.next;

    }
}