LeetCode刷题之Add Two Numbers
2018-04-25 本文已影响0人
JRTx
Problem
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.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
My Solution
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode p = l1, q = l2;
ListNode res = new ListNode(0);
ListNode n = res;
boolean flag = false;
while (p != null && q != null) {
int t = 0;
if (flag == true) {
t = p.val + q.val + 1;
} else {
t = p.val + q.val;
}
if (t < 10) {
n.next = new ListNode(t);
n = n.next;
flag = false;
} else {
n.next = new ListNode(t % 10);
n = n.next;
flag = true;
}
p = p.next;
q = q.next;
}
while (p != null) {
if (flag == true) {
int t = p.val + 1;
if (t < 10) {
n.next = new ListNode(t);
n = n.next;
flag = false;
} else {
n.next = new ListNode(t % 10);
n = n.next;
flag = true;
}
} else {
n.next = new ListNode(p.val);
n = n.next;
}
p = p.next;
}
while (q != null) {
if (flag == true) {
int t = q.val + 1;
if (t < 10) {
n.next = new ListNode(t);
n = n.next;
flag = false;
} else {
n.next = new ListNode(t % 10);
n = n.next;
flag = true;
}
} else {
n.next = new ListNode(q.val);
n = n.next;
}
q = q.next;
}
if (p == null && q == null && flag == true) {
n.next = new ListNode(1);
n = n.next;
flag = false;
} else {
}
return res.next;
}
}
Great Solution
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode p = l1, q = l2, curr = dummyHead;
int carry = 0;
while (p != null || q != null) {
int x = (p != null) ? p.val : 0;
int y = (q != null) ? q.val : 0;
int sum = carry + x + y;
carry = sum / 10;
curr.next = new ListNode(sum % 10);
curr = curr.next;
if (p != null) p = p.next;
if (q != null) q = q.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}
return dummyHead.next;
}