1290 Convert Binary Number in a Linked List to Integer 二进制链表转整数
Description:
Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.
Return the decimal value of the number in the linked list.
Example:
Example 1:
Input: head = [1,0,1]
Output: 5
Explanation: (101) in base 2 = (5) in base 10
Example 2:
Input: head = [0]
Output: 0
Example 3:
Input: head = [1]
Output: 1
Example 4:
Input: head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0]
Output: 18880
Example 5:
Input: head = [0,0]
Output: 0
Constraints:
The Linked List is not empty.
Number of nodes will not exceed 30.
Each node's value is either 0 or 1.
题目描述:
给你一个单链表的引用结点 head。链表中每个结点的值不是 0 就是 1。已知此链表是一个整数数字的二进制表示形式。
请你返回该链表所表示数字的 十进制值 。
示例 :
示例 1：
输入：head = [1,0,1]
输出：5
解释：二进制数 (101) 转化为十进制数 (5)
示例 2：
输入：head = [0]
输出：0
示例 3：
输入：head = [1]
输出：1
示例 4：
输入：head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0]
输出：18880
示例 5：
输入：head = [0,0]
输出：0
提示：
链表不为空。
链表的结点总数不超过 30。
每个结点的值不是 0 就是 1。
思路:
相当于按位将二进制转化为十进制
比如 1010 添上 1 -> 1010 * 2 + 1 -> 10100 + 1 -> 10100 | 1 -> 10101
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
public:
int getDecimalValue(ListNode* head)
{
int result = 0;
while (head)
{
result <<= 1;
result |= head -> val;
head = head -> next;
}
return result;
}
};
Java:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
private int level = 0;
public int getDecimalValue(ListNode head) {
if (head == null) return 0;
return getDecimalValue(head.next) + (int)(Math.pow(2, level++) * head.val);
}
}
Python:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def getDecimalValue(self, head: ListNode) -> int:
result = 0
while head:
result <<= 1
result |= head.val
head = head.next
return result
