1290. Convert Binary Number in a Linked List to Integer #
题目 #
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 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
or1
.
题目大意 #
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 。
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。
解题思路 #
- 给出一个链表,链表从头到尾表示的数是一个整数的二进制形式,要求输出这个整数的十进制。
- 简单题,从头到尾遍历一次链表,边遍历边累加二进制位。
代码 #
func getDecimalValue(head *ListNode) int {
sum := 0
for head != nil {
sum = sum*2 + head.Val
head = head.Next
}
return sum
}