Add Two Numbers
描述
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
分析
跟 Add Binary 很类似
代码
- Java
- C++
// Add Two Numbers
// 跟Add Binary 很类似
// 时间复杂度O(m+n)，空间复杂度O(1)
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1); // 头节点
int carry = 0;
ListNode prev = dummy;
for (ListNode pa = l1, pb = l2;
pa != null || pb != null;
pa = pa == null ? null : pa.next,
pb = pb == null ? null : pb.next,
prev = prev.next) {
final int ai = pa == null ? 0 : pa.val;
final int bi = pb == null ? 0 : pb.val;
final int value = (ai + bi + carry) % 10;
carry = (ai + bi + carry) / 10;
prev.next = new ListNode(value); // 尾插法
}
if (carry > 0)
prev.next = new ListNode(carry);
return dummy.next;
}
};
// Add Two Numbers
// 跟Add Binary 很类似
// 时间复杂度O(m+n)，空间复杂度O(1)
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode dummy(-1); // 头节点
int carry = 0;
ListNode *prev = &dummy;
for (ListNode *pa = l1, *pb = l2;
pa != nullptr || pb != nullptr;
pa = pa == nullptr ? nullptr : pa->next,
pb = pb == nullptr ? nullptr : pb->next,
prev = prev->next) {
const int ai = pa == nullptr ? 0 : pa->val;
const int bi = pb == nullptr ? 0 : pb->val;
const int value = (ai + bi + carry) % 10;
carry = (ai + bi + carry) / 10;
prev->next = new ListNode(value); // 尾插法
}
if (carry > 0)
prev->next = new ListNode(carry);
return dummy.next;
}
};