2017-03-12 00:15:42 +0000   |     algorithm leetcode linked list math   |   Viewed times   |    

题目

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.

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

注意

这题因为没有限定数字可以有多长,所以转型成int,long或者BigDecimal计算的方法都失败了。

朴素解法

短的那个数字,不足的位用0补足。

/**
 * 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) {
        ListNode result = new ListNode(-1); // HEAD
        ListNode index = result;
        int carry = 0;
        int num1 = 0;
        int num2 = 0;
        while (true) {
            if (l1 == null && l2 == null) {
                if (carry == 1) {
                    ListNode next = new ListNode(carry);
                    index.next = next;
                }
                return result.next;
            }
            if (l1 != null) {
                num1 = l1.val;
                l1 = l1.next;
            } else {
                num1 = 0;
            }
            if (l2 != null) {
                num2 = l2.val;
                l2 = l2.next;
            } else {
                num2 = 0;
            }
            int sum = num1 + num2 + carry;
            index.next = new ListNode(sum % 10);
            index = index.next;
            carry = sum / 10;
        }
    }
}

结果

add-two-numbers-1.png

优化两个数长度不相等的情况(递归版)

当一个数比另一个数更长,后半部分的加法就不用做,只是进了一位,或者直接照搬。

这个版本的缺点是,逻辑太复杂。

/**
 * 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.val == 0 && l1.next == null) { // base case l1 = 0
                return l2;
            }
            if (l2.val == 0 && l2.next == null) { // base case l2 = 0
                return l1;
            }
            ListNode result = new ListNode(-1); // HEAD is not included in the result
            ListNode index = result;
            int carry = 0;
            int num1 = 0;
            int num2 = 0;
            if (l2.val == 1 && l2.next == null) { // base case l2 = 1
                while (l1 != null && l1.val == 9) {
                    index.next = new ListNode(0);
                    l1 = l1.next;
                    index = index.next;
                    carry = 1;
                }
                if (l1 == null) {
                    index.next = new ListNode(1);
                } else {
                    ListNode newNode = new ListNode(l1.val + 1);
                    newNode.next = l1.next;
                    index.next = newNode;
                }
                return result.next;
            }
            while (true) {
                if (l1 == null && l2 == null) {
                    if (carry == 1) {
                        ListNode next = new ListNode(carry);
                        index.next = next;
                    }
                    return result.next;
                }
                if (l1 == null) { // l2更长,递归
                    index.next = addTwoNumbers(l2,new ListNode(carry));
                    return result.next;
                }
                if (l2 == null) { // l1更长,递归
                    index.next = addTwoNumbers(l1,new ListNode(carry));
                    return result.next;
                }
                // add two numbers on this bit, normal case
                int sum = l1.val + l2.val + carry;
                l1 = l1.next;
                l2 = l2.next;
                index.next = new ListNode(sum % 10);
                index = index.next;
                carry = sum / 10;
            }
    }
}

结果

虽然理论上当两个数字长度相差很大的时候,应该有所优化。但实际测试效果不理想,还不如朴素方法。 add-two-numbers-1.png