Add Two Numbers
- 问题: 给出两个链表分别表示倒序的非负整数,求两个整数的和
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null && l2 == null) {
return null;
}
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
int carry = 0;
ListNode curr = new ListNode(0);
ListNode dummy = curr;
while (l1 != null && l2 != null) {
curr.next = new ListNode((l1.val + l2.val + carry) % 10);
carry = (l1.val + l2.val + carry) / 10;
curr = curr.next;
l1 = l1.next;
l2 = l2.next;
}
while (l1 == null && l2 != null) {
curr.next = new ListNode((l2.val + carry) % 10);
carry = (l2.val + carry) / 10;
curr = curr.next;
l2 = l2.next;
}
while (l2 == null && l1 != null) {
curr.next = new ListNode((l1.val + carry) % 10);
carry = (l1.val + carry) / 10;
curr = curr.next;
l1 = l1.next;
}
if (carry != 0) {
curr.next = new ListNode(carry);
}
return dummy.next;
}
}
//优化while循环
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null && l2 == null) {
return null;
}
int carry = 0;
ListNode head = new ListNode(0);
ListNode dummy = head;
while (l1 != null || l2 != null) {
int sum = carry;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
head.next = new ListNode(sum % 10);
head = head.next;
carry = sum / 10;
}
if (carry != 0) {
head.next = new ListNode(carry);
}
return dummy.next;
}
}
445. Add Two Numbers II
- 问题: 给出两个链表分别表示正序的非负整数,求两个整数的和
- 思路: 先reverse然后再直接求加 -> 长度不等的时候, 需要考虑进位匹配问题 —> 计算长度用0补位 或者是先将长的跳过 从同一起点开始reverse, 高位是在前面,计算完了还要reverse回去