Java 程序添加两个链表表示的数字-集合 1
原文:https://www . geesforgeks . org/Java-program-to-add-two-numbers-由链表表示-set-1/
给定两个由两个列表表示的数字，编写一个返回求和列表的函数。求和列表是两个输入数字相加的列表表示。
例:
Input:
List1: 5->6->3 // represents number 563
List2: 8->4->2 // represents number 842
Output:
Resultant list: 1->4->0->5 // represents number 1405
Explanation: 563 + 842 = 1405
Input:
List1: 7->5->9->4->6 // represents number 75946
List2: 8->4 // represents number 84
Output:
Resultant list: 7->6->0->3->0// represents number 76030
Explanation: 75946+84=76030
接近:遍历两个列表，逐个选择两个列表的节点，并添加值。如果总和大于 10，则进位为 1 并减少总和。如果一个列表中的元素比另一个列表中的多，则认为该列表的剩余值为 0。
步骤为:
- 从头到尾遍历两个链表
- 将相应链接列表中的两位数字相加。
- 如果其中一个列表已到达末尾，则取 0 作为它的数字。
- 继续下去，直到两个列表都结束。
- 如果两位数之和大于 9，则将进位设置为 1，将当前数字设置为和% 10
下面是这个方法的实现。
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to add two numbers
// represented by linked list
class LinkedList
{
static Node head1, head2;
static class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Adds contents of two linked
lists and return the head node
of resultant list */
Node addTwoLists(Node first,
Node second)
{
// res is head node of the
// resultant list
Node res = null;
Node prev = null;
Node temp = null;
int carry = 0, sum;
// while both lists exist
while (first != null ||
second != null)
{
// Calculate value of next digit
// in resultant list. The next
// digit is sum of following things
// (i) Carry
// (ii) Next digit of first
// list (if there is a next digit)
// (ii) Next digit of second
// list (if there is a next digit)
sum = carry + (first != null ? first.data : 0) +
(second != null ? second.data : 0);
// Update carry for next calculation
carry = (sum >= 10) ? 1 : 0;
// Update sum if it is greater
// than 10
sum = sum % 10;
// Create a new node with sum as data
temp = new Node(sum);
// if this is the first node then set
// it as head of the resultant list
if (res == null)
{
res = temp;
}
// If this is not the first
// node then connect it to the rest.
else
{
prev.next = temp;
}
// Set prev for next insertion
prev = temp;
// Move first and second pointers
// to next nodes
if (first != null)
{
first = first.next;
}
if (second != null)
{
second = second.next;
}
}
if (carry > 0)
{
temp.next = new Node(carry);
}
// return head of the resultant
// list
return res;
}
/* Utility function to print a
linked list */
void printList(Node head)
{
while (head != null)
{
System.out.print(head.data +
" ");
head = head.next;
}
System.out.println("");
}
// Driver Code
public static void main(String[] args)
{
LinkedList list = new LinkedList();
// Creating first list
list.head1 = new Node(7);
list.head1.next = new Node(5);
list.head1.next.next = new Node(9);
list.head1.next.next.next =
new Node(4);
list.head1.next.next.next.next =
new Node(6);
System.out.print("First List is ");
list.printList(head1);
// Creating second list
list.head2 = new Node(8);
list.head2.next = new Node(4);
System.out.print("Second List is ");
list.printList(head2);
// Add the two lists and see the result
Node rs = list.addTwoLists(head1, head2);
System.out.print("Resultant List is ");
list.printList(rs);
}
}
// This code is contributed by Mayank Jaiswal
输出:
First List is 7 5 9 4 6
Second List is 8 4
Resultant list is 5 0 0 5 6
复杂度分析:
- 时间复杂度: O(m + n)，其中 m 和 n 分别是第一和第二列表中的节点数。 列表只需遍历一次。
- 空间复杂度: O(m + n)。 需要一个临时链表来存储输出号
更多详情请参考整篇文章添加两个链表表示的数字|集合 1 ！
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