題目2. 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
1
public class Solution {
private long getValue(ListNode list){
long value = 0;
int pow = 0;
while(list != null){
value = (long) (value+Math.pow(10, pow)*list.val);
list = list.next;
pow++;
}
return value;
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1 == null && l2 == null){
return null;
}
long result = getValue(l1) + getValue(l2);
if(result == 0){
return new ListNode(0);
}
ListNode head = new ListNode(-1);
ListNode tail = head;
while(result > 0){
ListNode temp = new ListNode((int) (result % 10));
tail.next = temp;
tail = temp;
result /= 10;
}
return head.next;
}
}
2,簡潔的寫法
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode prev = new ListNode(0);
ListNode head = prev;
int carry = 0;
while (l1 != null || l2 != null || carry != 0) {
ListNode cur = new ListNode(0);
int sum = ((l2 == null) ? 0 : l2.val) + ((l1 == null) ? 0 : l1.val) + carry;
cur.val = sum % 10;
carry = sum / 10;
prev.next = cur;
prev = cur;
l1 = (l1 == null) ? l1 : l1.next;
l2 = (l2 == null) ? l2 : l2.next;
}
return head.next;
}
