題目描述
思路
與989. 數(shù)組形式的整數(shù)加法的豎式計(jì)算方法基本一樣。無非是按位相加并且注意進(jìn)位及某個(gè)數(shù)組的數(shù)加完了的情況熬粗。
另外提一嘴爬坑,LeetCode的題目把好些邊界情況都給去了絮记,這樣不好栋艳。
Java代碼
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode res = new ListNode();
ListNode temp = new ListNode();
temp.next = res;
int carry = 0;
while (l1 != null && l2 != null) {
temp.next.val = l1.val + l2.val + carry;
if (temp.next.val >= 10) {
temp.next.val = temp.next.val % 10;
carry = 1;
} else {
carry = 0;
}
temp.next.next = new ListNode();
temp = temp.next;
l1 = l1.next;
l2 = l2.next;
}
if (l1 == null && l2 == null) {
if (carry == 1) {
temp.next.val = 1;
} else {
temp.next = null;
}
} else if (l1 == null && l2 != null) {
while (carry == 1 && l2 != null) {
temp.next.val = l2.val + carry;
if (temp.next.val >= 10) {
temp.next.val = temp.next.val % 10;
carry = 1;
} else {
carry = 0;
}
temp.next.next = new ListNode();
temp = temp.next;
l2 = l2.next;
}
if (l2 == null) {
if (carry == 1) {
temp.next.val = 1;
} else {
temp.next = null;
}
} else if (carry == 0) {
temp.next = l2;
}
} else if (l1 != null && l2 == null) {
while (carry == 1 && l1 != null) {
temp.next.val = l1.val + carry;
if (temp.next.val >= 10) {
temp.next.val = temp.next.val % 10;
carry = 1;
} else {
carry = 0;
}
temp.next.next = new ListNode();
temp = temp.next;
l1 = l1.next;
}
if (l1 == null) {
if (carry == 1) {
temp.next.val = 1;
} else {
temp.next = null;
}
} else if (carry == 0) {
temp.next = l1;
}
}
return res;
}
}
執(zhí)行用時(shí):2 ms, 在所有 Java 提交中擊敗了99.91%的用戶
內(nèi)存消耗:38.9 MB, 在所有 Java 提交中擊敗了30.50%的用戶