索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode
题目:https://oj.leetcode.com/problems/add-two-numbers/
代码(github):https://github.com/illuz/leetcode
求两个 List 相加产生的新的一个 List。
直接模拟就可以了。
C++:
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *ret = new ListNode(0);
ListNode *cur = ret;
int sum = 0;
while (1) {
if (l1 != NULL) {
sum += l1->val;
l1 = l1->next;
}
if (l2 != NULL) {
sum += l2->val;
l2 = l2->next;
}
cur->val = sum % 10;
sum /= 10;
if (l1 != NULL || l2 != NULL || sum)
cur = (cur->next = new ListNode(0));
else
break;
}
return ret;
}
};
Java:
public class Solution {
// Definition for singly-linked list.
public static class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode ret = new ListNode(0);
ListNode cur = ret;
int sum = 0;
while (true) {
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
cur.val = sum % 10;
sum /= 10;
if (l1 != null || l2 != null || sum != 0) {
cur = (cur.next = new ListNode(0));
} else {
break;
}
}
return ret;
}
}
Python:
class Solution:
# @return a ListNode
def addTwoNumbers(self, l1, l2):
ret = ListNode(0)
cur = ret
sum = 0
while True:
if l1 != None:
sum += l1.val
l1 = l1.next
if l2 != None:
sum += l2.val
l2 = l2.next
cur.val = sum % 10
sum /= 10
if l1 != None or l2 != None or sum != 0:
cur.next = ListNode(0)
cur = cur.next
else:
break
return ret