题目描述
原题
Description:
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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
原题翻译
描述:
给定两个非空链表,表示两个非负整数。数字以相反的顺序存储,每个节点包含一个数字。将两个数字相加并将其作为链表返回。可以认为这两个数字不包含任何前导0,除了数字0本身。
例如:
输入: (2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 0 -> 8
解释: 342 + 465 = 807.
解法一(mine)
主要思想
将两个数字表示出来,相加,再转化成结果链表
(考虑到两数和可能超过了int的限制,选择BigInteger)
想法很简单,然而…Memory Limit Exceeded(内存超过限制)
源码
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode ln1 = l1, ln2 = l2; // 不破坏原链表
BigInteger a, b, temp1, temp2, b10;
b10 = new BigInteger("10");
temp1 = temp2 = new BigInteger("1");
a = b = new BigInteger("0");
while(ln1 != null) {
a = a.add(new BigInteger(ln1.val + "").multiply(temp1));
ln1 = ln1.next;
temp1 = temp1.multiply(b10);
}
while(ln2 != null) {
b = b.add(new BigInteger(ln2.val + "").multiply(temp2));
ln2 = ln2.next;
temp2 = temp2.multiply(b10);
}
BigInteger result = a.add(b);
ListNode res;
ListNode t = new ListNode(result.mod(b10).intValue());
result = result.divide(b10);
res = t;
while(result != new BigInteger("0")) {
t.next = new ListNode(result.mod(b10).intValue());
t = t.next;
result = result.divide(b10);
}
return res;
}
}
解法二
主要思想
同时遍历两链表,将对应位上对两数相加(超过10的部分除以10再存起来,下一位继续用)
时间复杂度:O(n)
源码
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode ln1 = l1, ln2 = l2, head = null, node = null;
int carry = 0, remainder = 0, sum = 0;
head = node = new ListNode(-1);
// 遍历l1和l2
while(ln1 != null || ln2 != null || carry != 0) {
sum = (ln1 != null ? ln1.val : 0) + (ln2 != null ? ln2.val : 0) + carry;
carry = sum / 10;
remainder = sum % 10;
node = node.next = new ListNode(remainder);
ln1 = (ln1 != null ? ln1.next : null);
ln2 = (ln2 != null ? ln2.next : null);
}
return head.next;
}
}
这个解法超过了57.12%的答案。虽然对于计算机来说不是最佳答案,但对于人来说它是易读的
解法三
第一名答案
主要思想
与解法二的思想类似。
源码
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);
int carry = 0;
while (l1 != null || l2 != null || carry > 0) {
ListNode itr = head;
while (itr.next != null) {
itr = itr.next;
}
int sum = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry);
carry = sum / 10;
ListNode temp = new ListNode(sum % 10);
itr.next = temp;
if (l1 != null)
l1 = l1.next;
if (l2 != null)
l2 = l2.next;
}
return head.next;
}
}