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package question0002_add_two_numbers;
/**
* 递归实现。
*
* 时间复杂度和空间复杂度均是O(n1 + n2)，其中n1为链表l1的长度，n2位链表l2的长度。
*
* 执行用时：2ms，击败99.95%。消耗内存：39.9MB，击败95.88%。
*/
public class Solution3 {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
return addTwoNumbers(l1, l2, 0);
}
private ListNode addTwoNumbers(ListNode l1, ListNode l2, int flag) {
if (l1 == null && l2 == null) {
if (flag == 0) {
return null;
}
return new ListNode(1);
} else if (l1 == null) {
if (flag == 0) {
return l2;
}
int num = l2.val + flag;
if (num >= 10) {
num -= 10;
flag = 1;
} else {
flag = 0;
}
l2.val = num;
l2.next = addTwoNumbers(l1, l2.next, flag);
return l2;
} else if (l2 == null) {
if (flag == 0) {
return l1;
}
int num = l1.val + flag;
if (num >= 10) {
num -= 10;
flag = 1;
} else {
flag = 0;
}
l1.val = num;
l1.next = addTwoNumbers(l1.next, l2, flag);
return l1;
}
int num = l1.val + l2.val + flag;
if (num >= 10) {
num -= 10;
flag = 1;
} else {
flag = 0;
}
l1.val = num;
l1.next = addTwoNumbers(l1.next, l2.next, flag);
return l1;
}
}